Class 12 Maths Chapter 7 Miscellaneous Exercise

Given: \( \frac{1}{x-x^{3}} \)
Let \( \mathrm{I}=\frac{1}{x-x^{3}}=\frac{1}{x\left(1-x^{2}\right)}=\frac{1}{x(1+x)(1-x)} \)
Using partial differentiation:
Let \( \frac{1}{x(1+x)(1-x)}=\frac{A}{x}+\frac{B}{1+x}+\frac{C}{1-x} \quad \quad \ldots\text{(1)}\)
\(\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A(1+x)(1-x)+B(x)(1-x)+C(x)(1+x)}{x(1+x)(1-x)}\)
\(\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{A\left(1+x^{2}\right)+B x(1-x)+C x(1+x)}{x(1+x)(1-x)}\)
\(\Rightarrow 1=\mathrm{A}-\mathrm{Ax}^{2}+\mathrm{B}x-\mathrm{B}x^{2}+\mathrm{C}x+\mathrm{C}x^{2}\)
\( \Rightarrow 1=\mathrm{A}+(\mathrm{B}+\mathrm{C}) x+(-\mathrm{A}-\mathrm{B}+\mathrm{C}) x^{2} \)
Equating the coefficients of \( x, x^{2} \) and constant value. We get:
(a) \( \mathrm{A}=1 \)
(b) \( \mathrm{B}+\mathrm{C}=0 \)
\(\Rightarrow \mathrm{B}=-\mathrm{C} \)
(c) \( -\mathrm{A}-\mathrm{B}+\mathrm{C}=0 \)
\(\Rightarrow-1-(-C)+C=0\)
\(\Rightarrow 2 \mathrm{C}=1\)
\( \Rightarrow \mathrm{C}=\frac{ 1 }{ 2 }\)
So, \( B=-\frac{ 1 }{ 2 } \)
Put these values in equation (1)
\(\Rightarrow \frac{1}{x(1+x)(1-x)}=\frac{1}{x}+\frac{-\left(\frac{1}{2}\right)}{1+x}+\frac{\left(\frac{1}{2}\right)}{1-x}\)
\(\Rightarrow \int \frac{1}{x(1+x)(1-x)} d x=\int \frac{1}{x} d x-\frac{1}{2} \int \frac{1}{1+x} d x+\frac{1}{2} \int \frac{1}{1-x} d x\)
\(=\log |x|-\frac{1}{2} \log |1+x|+\frac{1}{2} \log |1-x|\)
\(=\log |x|-\log \left|(1+x)^{\frac{1}{2}}\right|+\log \left|(1-x)^{\frac{1}{2}}\right|\)
\(=\log \left|\frac{x}{(1+x)^{\frac{1}{2}}(1-x)^{\frac{1}{2}}}\right|+C\)
\(=\log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{(1+x)(1-x)^{^{\frac{1}{2}}}}\right|+C\)
\(=\log \left|\frac{\left(x^{2}\right)^{\frac{1}{2}}}{\left(1-x^{2}\right)^{\frac{1}{2}}}\right|+C\)
\(=\log \left|\left(\frac{x^{2}}{1-x^{2}}\right)^{\frac{1}{2}}\right|+C\)
\(\Rightarrow \mathrm{I}=\frac{1}{2} \log \left|\frac{x^{2}}{1-x^{2}}\right|+C\)

Given: \( \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \)
Let \( \mathrm{I}=\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \)
Multiply and divide by, \( \sqrt{x+a}-\sqrt{x+b} \)
\(\Rightarrow \mathrm{I}=\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}\)
\(=\frac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a})^{2}-(\sqrt{x+b})^{2}}\)
\(=\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}\)
\(=\frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}\)
\(=\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x=\int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b} d x\)
\(=\frac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x\)
\(=\frac{1}{a-b} \int\left((x+a)^{\frac{1}{2}}(x+b)^{\frac{1}{2}}\right) d x\)
\(=\frac{1}{a-b}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}\right]\)
\(\Rightarrow \mathrm{I}=\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+C\)

Given: \( \frac{1}{x \sqrt{a x-x^{2}}} \)
Let \( \mathrm{I}=\frac{1}{x \sqrt{a x-x^{2}}} \)
Put \( x=\frac{a}{t} \)
\(\Rightarrow d x=-\frac{a}{t^{2}} d t \)
\(\Rightarrow \int \frac{1}{x \sqrt{a x-x^{2}}} d x=\int \frac{1}{\frac{a}{t} \sqrt{\frac{a \cdot a}{t}}-\left(\frac{a}{t}\right)^{2}} \cdot-\frac{a}{t^{2}} d t\)
\(=\int \frac{-1}{a t} \cdot \frac{1}{\sqrt{\frac{1}{t}-\left(\frac{1}{t}\right)^{2}}} d t\)
\(=\frac{-1}{a t} \int \frac{1}{\sqrt{\frac{1}{t}-\left(\frac{1}{t}\right)^{2}}} d t\)
\(=-\frac{1}{a} \int \frac{1}{\sqrt{t-1}} d t\)
\(=-\frac{1}{a} \int(t-1)^{-\frac{1}{2}} d t\)
\(=-\frac{1}{a}\left[\frac{\sqrt{t-1}}{\frac{1}{2}}\right]+C\)
\(=-\frac{2}{a}\left[\sqrt{\left(\frac{a}{x}-1\right)}\right]+C \text { because, } \mathrm{t}=\frac{a}{x}\)
\(=\mathrm{I}=-\frac{2}{a}\left[\sqrt{\left(\frac{a}{x}-1\right)}\right]+C\)

Given: \( \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}} \)
Let \( \mathrm{I}=\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}} \)
Multiply and divide by \( x^{-3} \), we get
\(\frac{x^{-3}}{x^{2} \cdot x^{-3}\left(x^{4}+1\right)^{\frac{3}{4}}}=\frac{x^{-3} \cdot\left(x^{4}+1\right)^{\frac{3}{4}}}{x^{2} \cdot x^{-3}}\)
\(=\frac{\left(x^{4}+1\right)^{-\frac{3}{4}}}{x^{5} \cdot x^{-3 \times \frac{4}{4}}}\)
\(=\frac{\left(x^{4}+1\right)^{-\frac{3}{4}}}{x^{5} \cdot\left(x^{4}\right)^{-\frac{3}{4}}}\)
\(=\frac{1}{x^{5}} \cdot\left(\frac{x^{4}+1}{x^{4}}\right)^{-\frac{3}{4}}\)
Let \( \frac{1}{x^{4}}=t=(x)^{-4} \)
\(\Rightarrow \frac{-4}{x^{5}} d x=d t \)
\(\Rightarrow \frac{1}{x^{5}} d x=-\frac{d t}{4} \)
\(\Rightarrow \int \frac{1}{x^{2} \cdot\left(x^{4}+1\right)^{-\frac{3}{4}}} d x=\int \frac{1}{x^{5}} \cdot\left(1+\frac{1}{x^{4}}\right)^{-\frac{3}{4}} \cdot d x\)
\(=\int(1+t)^{-\frac{3}{4}} \cdot\left(-\frac{d t}{4}\right)\)
\(=-\frac{1}{4} \int(1+t)^{-\frac{3}{4}} \cdot d t\)
\(=-\frac{1}{4}\left[\frac{(1+t)^{\frac{1}{4}}}{\frac{1}{4}}\right]+C\)
\(=-\frac{1}{4}\left[\frac{\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}}{\frac{1}{4}}\right]+C\)
\(=-\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+C\)

Given: \( \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} \)
or we can write it as, \( \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{2}}\left(1+x^{\frac{1}{6}}\right)} \)
Let \( x=t^{6} \)
\(\Rightarrow d x=6 t^{5} d t \)
\(\Rightarrow \int \frac{1}{x^{\frac{1}{2}}\left(1+x^{\frac{1}{6}}\right)} \cdot d x=\int \frac{6 t^{5}}{t^{2}(1+t)} \cdot d t\)
\(=6 \cdot \int \frac{t^{3}}{(1+t)} \cdot d t\)
After division we get,
\(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=6 \cdot \int\left[\left(t^{2}-t+1\right)-\frac{1}{(1+t)}\right] \cdot d t\)
\(=6 \cdot\left\{\int t^{2} \cdot d t-\int t \cdot d t+\int 1 \cdot d t-\int\left[\frac{1}{(1+t)}\right] \cdot d t\right\}\)
\(=6\left[\left(\frac{t^{3}}{3}\right)-\left(\frac{t^{2}}{2}\right)+t-\log (1+t)\right]\)
\(=6\left[\left(\frac{\left(x^{\frac{1}{6}}\right)^{3}}{3}\right)-\left(\frac{\left(x^{\frac{1}{6}}\right)^{2}}{2}\right)+\left(x^{\frac{1}{6}}\right)-\log \right.\)\(\left.\left(1+\left(x^{\frac{1}{6}}\right)\right)\right]+C\)
\(=\left[\left(2 x^{\frac{1}{2}}\right)-\left(3 x^{\frac{1}{3}}\right)+6 \cdot x^{\frac{1}{6}}-6 \cdot \log \left(1+x^{\frac{1}{6}}\right)\right]+C\)
\(=2 \sqrt{x}-\left(3 x^{\frac{1}{3}}\right)+6 x^{\frac{1}{6}}-6 \cdot \log \left(1+x^{\frac{1}{6}}\right)+C\)

Given: \( \frac{5 x}{(x+1)\left(x^{2}+9\right)} \)
Let \( \mathrm{I}=\frac{5 x}{(x+1)\left(x^{2}+9\right)} \quad \quad \ldots\text{(1)}\)
Using partial differentiation:
Let \( \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)} \)
\(\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A\left(x^{2}+9\right)+(B x+C)(x+1)}{(x+1)\left(x^{2}+9\right)}\)
\(\Rightarrow 5 =\mathrm{A}\left(x^{2}+9\right)+(\mathrm{B}x+\mathrm{C})(+1)\)
\(\Rightarrow 5 =\mathrm{A}x^{2}+9 \mathrm{A}+\mathrm{B}x^{2}+\mathrm{B}x+\mathrm{C}x+\mathrm{C}\)
\(\Rightarrow 5 =9 \mathrm{A}+\mathrm{C}+(\mathrm{B}+\mathrm{C}) x+(\mathrm{A}+\mathrm{B}) x^{2}\)
Equating the coefficients of \( x, x^{2} \) and constant value. We get:
(a) \( 9 \mathrm{A}+\mathrm{C}=0 \)
\(\Rightarrow \mathrm{C}=-9 \mathrm{A} \)
(b) \( \mathrm{B}+\mathrm{C}=5\)
\( \Rightarrow \mathrm{B}=5-\mathrm{C} \)
\(\Rightarrow \mathrm{B}=5-(-9 \mathrm{~A}) \)
\(\Rightarrow \mathrm{B}=5+9 \mathrm{A} \)
(c) \( \mathrm{A}+\mathrm{B}=0 \)
\(\Rightarrow \mathrm{A}=-\mathrm{B} \)
\(\Rightarrow \mathrm{A}=-(5+9 \mathrm{A}) \)
\(\Rightarrow 10 \mathrm{A}=-5 \)
\(\Rightarrow \mathrm{A}=-\frac{ 1 }{ 2 } \)
and \( C=\frac{ 9 }{ 2 } \) and \( B=\frac{ 1 }{ 2 } \)
Put these values in equation (1)
\(\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)}\)
\(\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{-\frac{1}{2}}{(x+1)}+\frac{\left(\frac{1}{2}\right) x+\frac{9}{2}}{\left(x^{2}+9\right)}\)
\(\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=-\frac{1}{2} \cdot \frac{1}{(x+1)}+\frac{1}{2} \frac{x+9}{\left(x^{2}+9\right)}\)
\(\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x=-\frac{1}{2} \cdot \int \frac{1}{(x+1)} d x+\frac{1}{2} \cdot \int \frac{x}{\left(x^{2}+9\right)} \)\(d x+\frac{9}{2} \int \frac{1}{\left(x^{2}+9\right)} d x\)
\(\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x=-\frac{1}{2} \cdot \int \frac{1}{(x+1)} d x+\mathrm{I}_{1}+\)\(\frac{9}{2} \int \frac{1}{\left(x^{2}+\left(3^{2}\right)\right)} d x\)
\(\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x=-\frac{1}{2} \cdot \log |x+1|+I_{1}+\)\(\frac{9}{2} \cdot\left(\frac{1}{3} \tan ^{-1} \frac{x}{3}\right) \quad \ldots\text{(2)}\)
First solve for \( \mathrm{I}_{1} \) :
\(\Rightarrow \mathrm{I}_{1}=\frac{1}{2} \cdot \int \frac{1}{(t+9)} \cdot \frac{d t}{2}\)
\(\Rightarrow \mathrm{I}_{1}=\frac{1}{4} \log |t+9|\)
\(\Rightarrow \mathrm{I}_{1}=\frac{1}{4} \log \left|x^{2}+9\right|\)
Put the value in equ. (2)
\(\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x=-\frac{1}{2} \cdot \log |x+1|+\frac{1}{4} \log \left|x^{2}+9\right|+\)\(\frac{3}{2} \cdot\left(\frac{1}{3} \tan ^{-1} \frac{x}{3}\right)+C\)

Given: \( \frac{\sin x}{\sin (x-a)} \)
Let \( \mathrm{I}=\frac{\sin x}{\sin (x-a)} \)
Let \(x -\mathrm{a}=\mathrm{t} \)
\(\Rightarrow x=\mathrm{t}+\mathrm{a} \)
\(\Rightarrow \mathrm{d}x=\mathrm{dt} \)
\(\Rightarrow \int \frac{\sin x}{\sin (x-a)} d x=\int \frac{\sin (t+a)}{\sin (t)} d t\)
As, \( \{\sin (A+B)=\sin A \cos B+\cos A \sin B\} \)
\(\Rightarrow \int \frac{\sin x}{\sin (x-a)} d x=\int \frac{\sin t \cos a+\cos t \sin a}{\sin (t)} d t\)
\(=\int \frac{\sin t \cos a}{\sin t}+\frac{\cos t \sin a}{\sin t} d t\)
\(=\int(\cos a+\cot t \sin a) d t\)
\(=(\cos a) \cdot \int 1 \cdot d t+\sin a \cdot \int(\cot t) d t\)
\(=(\cos a) \cdot t+\sin a \cdot \log |\sin t|+C\)
\(=(\cos a) \cdot(x-a)+\sin a \cdot \log |\sin (x-a)|+C\)
\(=\sin a \cdot \log |\sin (x-a)|+x \cdot \cos a-a \cdot \cos a+C\)
\(=\sin a \cdot \log |\sin (x-a)|+x \cdot \cos a+C_{2}\)

Given: \( \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} \)
Let \( \mathrm{I}=\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} \)
\(\Rightarrow \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}=\frac{e^{5 \log x}\left(e^{\log x}-1\right)}{e^{3 \log x}\left(e^{\log x}-1\right)}\)
\(=e^{2 \log x}\)
\(=e^{\log x}\)
\(=x^{2}\)
\(\Rightarrow \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=\int x^{2} d x\)
\(=\frac{x^{3}}{3}+C\)
\(\Rightarrow \mathrm{I}=\frac{x^{3}}{3}+C\)

Given: \( \frac{\cos x}{\sqrt{4-\sin ^{2} x}} \)
Let \( \mathrm{I}=\frac{\cos x}{\sqrt{4-\sin ^{2} x}} \)
Put \( \sin x=\mathrm{t} \Rightarrow \cos x\mathrm{d}x=\mathrm{dt} \)
\(\Rightarrow \int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x=\int \frac{1}{\sqrt{4-t^{2}}} d t\)
\(=\int \frac{1}{\sqrt{\left((2)^{2}-t^{2}\right)}} d t\)
\(=\sin ^{-1}\left(\frac{t}{2}\right)+C\)
\(\Rightarrow \mathrm{I}=\sin ^{-1}\left(\frac{\sin x}{2}\right)+C\)

Given: \( \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cdot \cos ^{2} x} \)
Let \( \mathrm{I}=\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cdot \cos ^{2} x} \)
\(\Rightarrow \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cdot \cos ^{2} x}=\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{4} x-\cos ^{4} x\right)}{\sin ^{2} x+\cos ^{2} x-\sin ^{2} x \cdot \cos ^{2} x-\sin ^{2} x \cdot \cos ^{2} x}\)
\(=\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)\left(\sin ^{2} x+\cos ^{2} x\right)}{\left(\sin ^{2} x-\sin ^{2} x \cdot \cos ^{2} x\right)+\left(\cos ^{2} x-\sin ^{2} x \cdot \cos ^{2} x\right)}\)
\(=\frac{\left(\sin ^{4} x+\cos ^{4} x\right)\left(\sin ^{2} x-\cos ^{2} x\right) \cdot(1)}{\sin ^{2} x\left(1-\cos ^{2} x\right)+\cos ^{2} x\left(1-\sin ^{2} x\right)}\)
\(=\frac{-\left(\sin ^{4} x+\cos ^{4} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)}{\sin ^{2} x\left(\sin ^{2} x\right)+\cos ^{2} x\left(\cos ^{2} x\right)}\)
\(=\frac{-\left(\sin ^{4} x+\cos ^{4} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)}{\left(\sin ^{4} x+\cos ^{4} x\right)}\)
\(=\left(\sin ^{4} x+\cos ^{4} x\right)\)
\(=-\cos 2 x\)

Given: \( \frac{1}{\cos (x+a) \cos (x+b)} \)
Let \( \mathrm{I}=\frac{1}{\cos (x+a) \cos (x+b)} \)
Multiply and divide by \( \sin (\mathrm{a}-\mathrm{b}) \), we get
\(\mathrm{I}=\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (a-b)}{\cos (x+a) \cos (x+b)}\right)\)
\(=\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (a-b+x-x)}{\cos (x+a) \cos (x+b)}\right)\)
\(=\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)}\right)\)
As, \( \{\sin (A-B)=\sin A \cos B-\cos A \sin B\} \)
\(\Rightarrow \mathrm{I}=\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a) \cdot \cos (x+b)-\cos (x+a) \cdot \sin (x+b)}{\cos (x+a) \cos (x+b)}\right)\)
\(=\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a) \cdot \cos (x+b)}{\cos (x+a) \cos (x+b)}-\frac{\cos (x+a) \cdot \sin (x+b)}{\cos (x+a) \cos (x+b)}\right)\)
\(=\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a)}{\cos (x+a)}-\frac{\sin (x+b)}{\cos (x+b)}\right)\)
\(=\frac{1}{\sin (a-b)} \cdot[\tan (x+a)-\tan (x+b)]\)
\(= \int \frac{1}{\cos (x+a) \cos (x+b)} d x=\int \frac{1}{\sin (a-b)} \cdot[\tan (x+a)\)\(-\tan (x+b)]\)
\(=\frac{1}{\sin (a-b)} \cdot\left\{\int \tan (x+a) d x-\int \tan (x+b) d x\right\}\)
\(=\frac{1}{\sin (a-b)}[-\log |\cos (x+a)|+(-\log |\cos (x+a)|)]\)
\(=\frac{1}{\sin (a-b)}[-\log |\cos (x+a)+\log | \cos (x+a) \mid]\)
\(= \mathrm{I}=\frac{1}{\sin (a-b)} \cdot \log \left|\frac{\cos (x+b)}{\cos (x+a)}\right|+C\)

Given: \( \frac{x^{3}}{\sqrt{1-x^{8}}} \)
Let \( \mathrm{I}=\frac{x^{3}}{\sqrt{1-x^{8}}} \)
Now, let \( x^{4}=\mathrm{t} \)
\(\Rightarrow 4 ^{3} \mathrm{d}x=\mathrm{dt} \)
And \(x ^{3} \mathrm{d}x=\frac{ \mathrm{dt} }{ 4 } \)
\(\Rightarrow \int \frac{x^{3}}{\sqrt{1-x^{8}}} d x=\int \frac{1}{\sqrt{1-t^{2}}}\left(\frac{d t}{4}\right)\)
\(=\frac{1}{4} \int \frac{1}{\sqrt{1-t^{2}}} \cdot d t\)
\(=\frac{1}{4} \sin ^{-1} t+C\)
\(\Rightarrow \mathrm{I}=\frac{1}{4} \sin ^{-1}\left(x^{4}\right)+C\)

Given: \( \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} \)
Let \( \mathrm{I}=\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} \)
Let \( \mathrm{e}^{x}=\mathrm{t} \Rightarrow \mathrm{e}^{x} \mathrm{d}x=\mathrm{dt} \)
\(\Rightarrow \int \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} d x=\int \frac{1}{(1+t)(2+t)} d t\)
\(=\int\left[\frac{1}{(1+t)}-\frac{1}{(2+t)}\right] d t\)
\(=\int\left[\frac{1}{(1+t)}\right] d t-\int\left[\frac{1}{(2+t)}\right] d t\)
\(=\log |(1+t)|-\log |(2+t)|+C\)
\(=\log \left|\frac{1+t}{2+t}\right|+C\)
\(\Rightarrow \mathrm{I}=\log \left|\frac{1+e^{x}}{2+e^{x}}\right|+C\)

Given: \( \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} \)
Let \( \mathrm{I}=\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} \)
Using partial differentiation:
Let \( \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}\quad\ldots \text{(1)} \)
\(\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+9\right)}=\frac{(A x+B)\left(x^{2}+4\right)+(C x+D)+\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{2}+9\right)}\)
\(\Rightarrow 1=(\mathrm{A}x+\mathrm{B})\left(x^{2}+4\right)+(\mathrm{C}x+\mathrm{D})\left(^{2}+1\right)\)
\(\Rightarrow 1=\mathrm{A}x^{3}+4 \mathrm{A}x+\mathrm{B}x^{2}+4 \mathrm{B}+\mathrm{C}x^{3}+\)\(\mathrm{C}x+\mathrm{D}x^{2}+\mathrm{D}\)
\(\Rightarrow 1=(\mathrm{A}+\mathrm{C}) x^{3}+(\mathrm{B}+\mathrm{D}) ^{2}+(4 \mathrm{A}+\mathrm{C}) +\)\((4 \mathrm{B}+\mathrm{D})\)
Equating the coefficients of \( x, x^{2}, x^{3} \) and constant value. We get:
(a) \( \mathrm{A}+\mathrm{C}=0 \)
\(\Rightarrow \mathrm{C}=-\mathrm{A} \)
(b) \( \mathrm{B}+\mathrm{D}=0 \)
\(\Rightarrow \mathrm{B}=-\mathrm{D} \)
(c) \( 4 \mathrm{~A}+\mathrm{C}=0 \)
\(\Rightarrow 4 \mathrm{A}=-\mathrm{C} \)
\(\Rightarrow 4 \mathrm{A}=\mathrm{A} \)
\(\Rightarrow 3 \mathrm{A}=0 \)
\(\Rightarrow \mathrm{A}=0 \)
\(\Rightarrow \mathrm{C}=0 \)
(d) \( 4 \mathrm{B}+\mathrm{D}=1 \)
\(\Rightarrow 4 \mathrm{B}-\mathrm{B}=1 \)
\(\Rightarrow \mathrm{B}=\frac{ 1 }{ 3 } \)
\(\Rightarrow \mathrm{D}=-\frac{ 1 }{ 3 } \)
Put these values in equation (1)
\(\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}\)
\(\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{(0) x+\frac{1}{3}}{\left(x^{2}+1\right)}+\frac{(0) x+\left(-\frac{1}{3}\right)}{\left(x^{2}+4\right)}\)
\(\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{\frac{1}{3}}{\left(x^{2}+1\right)}+\frac{\left(-\frac{1}{3}\right)}{\left(x^{2}+4\right)}\)
\(\Rightarrow \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+1\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+4\right)} d x\)
\(\Rightarrow \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+1^{2}\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+2^{2}\right)} d x\)
\(=\frac{1}{3} \cdot \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C\)
\(\Rightarrow \mathrm{I}=\frac{1}{3} \cdot \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C\)

Given: \( \cos ^{3} x e^{\log \sin x} \)
Let \( \mathrm{I}=\cos ^{3} x e^{\log \sin x} \)
\(=\cos ^{3} x \cdot \sin x\)
Let \( \cos =\mathrm{t} \Rightarrow-\sin x\mathrm{d}x=\mathrm{dt} \Rightarrow \sin x\mathrm{d}x=\mathrm{dt} \)
\(\Rightarrow \int \cos ^{3} x e^{\log \sin x} d x=\int \cos ^{3} x \cdot \sin x d x\)
\(=\int t^{3},(-d t)\)
\(=-\int t^{3} \cdot d t\)
\(=-\frac{t^{4}}{4}+C\)
\(=-\frac{\cos ^{4} x}{4}+C\)

Given: \( e^{3 \log x}\left(x^{4}+1\right)^{-1} \)
Let \( \mathrm{I}=e^{3 \log x}\left(x^{4}+1\right)^{-1} \)
\(=e^{3 \log x}\left(x^{4}+1\right)^{-1}\)
\(=\frac{x^{3}}{x^{4}+1}\)
Let \( x^{4}=\mathrm{t} \Rightarrow 4 x^{3} \mathrm{d}x=\mathrm{dt} \Rightarrow x^{3} \mathrm{d}x=\frac{ \mathrm{dt} }{ 4 } \)
\(\Rightarrow \int e^{3 \log x}\left(x^{4}+1\right)^{-1}=\int \frac{x^{3}}{x^{4}+1} d x\)
\(=\int \frac{1}{t+1} \cdot \frac{d t}{4}\)
\(=\frac{1}{4} \cdot \int \frac{1}{t+1} \cdot d t\)
\(=\frac{1}{4} \log (t+1)+C\)
\(\Rightarrow \mathrm{I}=\frac{1}{4} \log \left(x^{4}+1\right)+C\)

Given: \( \mathrm{f}^{\prime}(\mathrm{a}x+\mathrm{b})[\mathrm{f}(\mathrm{a}x+\mathrm{b})]^{n} \)
Let \( \mathrm{f}(\mathrm{a}x+\mathrm{b})=\mathrm{t} \)
\(\Rightarrow \mathrm{a} . \mathrm{f}^{\prime}(\mathrm{a}x+\mathrm{b}) \mathrm{d}x=\mathrm{dt} \)
\(\Rightarrow \int \mathrm{f}^{\prime}(\mathrm{a}x+\mathrm{b})[\mathrm{f}(\mathrm{a}x+\mathrm{b})]^{n}=\int t^{n}\left(\frac{d t}{a}\right)\)
\(=\frac{1}{a} \int t^{n} d t\)
\(=\frac{1}{a} \cdot \frac{t^{n+1}}{n+1}+C\)
\(=\frac{1}{a} \cdot \frac{(\mathrm{f}(\mathrm{a}x+\mathrm{b}))^{n+1}}{n+1}+C\)
\(=\frac{1}{a(n+1)} \cdot(\mathrm{f}(\mathrm{a}x+\mathrm{b}))^{n+1}+C\)

Given: \( \frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} \)
Let \( \mathrm{I}=\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} \)
As, \( \{\sin (A+B)=\sin A \cos B+\cos A \sin B\} \)
\(\Rightarrow \mathrm{I}=\frac{1}{\sqrt{\sin ^{3} x(\sin x \cos \alpha+\cos x \sin \alpha)}}\)
\(\Rightarrow \mathrm{I}=\frac{1}{\sqrt{\sin ^{3} x\left(\sin x \cos \alpha+\cos x \cdot \frac{\sin x}{\sin x} \sin \alpha\right)}}\)
\(=\frac{1}{\sqrt{\sin ^{3} x\left(\sin x \cos \alpha+\cos x \cdot \frac{\cos x}{\sin x} \sin \alpha\right)}}\)
\(=\frac{1}{\sqrt{\sin ^{3} x(\cos \alpha+\cot x \sin \alpha)}}\)
\(=\frac{1}{\sin ^{3} x \sqrt{(\cos \alpha+\cot x \sin \alpha)}}\)
\(=\frac{\operatorname{cosec}^{2} x}{\sqrt{(\cos \alpha+\cot x \sin \alpha)}}\)
Now, let \( (\cos \alpha+\cot x \sin \alpha)=t \)
\(\Rightarrow-\operatorname{cosec}^{2} x \cdot \sin \alpha d x=d t \)
\(=\int \frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} d x=\int \frac{\operatorname{cosec}^{2} x}{\sqrt{(\cos \alpha+\cot x \sin \alpha)}} d x\)
\(=\int \frac{1}{\sqrt{t}} \cdot-\frac{d t}{\sin \alpha}\)
\(=-\frac{1}{\sin \alpha} \int \frac{1}{\sqrt{t}} \cdot d t\)
\(=-\frac{1}{\sin \alpha} \int t^{-\frac{1}{2}} \cdot d t\)
\(=-\frac{1}{\sin \alpha}\left[\frac{t \frac{1}{2}}{\frac{1}{2}}\right]+C\)
\(=-\frac{2}{\sin \alpha}[\sqrt{t}]+C\)
\(=-\frac{2}{\sin \alpha}[\sqrt{(\cos \alpha+\cot x \sin \alpha)}]+C\)
\(=-\frac{2}{\sin \alpha}\left[\sqrt{\left(\cos \alpha+\frac{\cos x}{\sin x} \sin \alpha\right)}\right]+C\)
\(=-\frac{2}{\sin \alpha}\left[\sqrt{\frac{(\cos \alpha \sin x+\cos x \sin \alpha)}{\sin x}}\right]+C\)
\(\Rightarrow \mathrm{I}=-\frac{2}{\sin \alpha}\left[\sqrt{\frac{\sin (x+\alpha)}{\sin x}}\right]+C\)

Given: \( \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}} \)
Let \( \mathrm{I}=\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}} \quad \quad \ldots\text{(1)}\)
As we know, \( \sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}=\frac{\pi}{2} \)
\(\Rightarrow \mathrm{I}=\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}=\frac{\left(\frac{\pi}{2}-\cos ^{-1} \sqrt{x}\right)-\cos ^{-1} \sqrt{x}}{\left(\frac{\pi}{2}\right)}\)
\(\Rightarrow \int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}} d x=\int \frac{\left(\frac{\pi}{2}-\cos ^{-1} \sqrt{x}\right)-\cos ^{-1} \sqrt{x}}{\left(\frac{\pi}{2}\right)} d x\)
\(=\left(\frac{2}{\pi}\right) \int\left(\frac{\pi}{2}-\cos ^{-1} \sqrt{x}\right)-\cos ^{-1} \sqrt{x} d x\)
\(=\left(\frac{2}{\pi}\right) \int\left(\frac{\pi}{2} \cdot d x\right)-\left(\frac{2}{\pi}\right) \int\left(\cos ^{-1} \sqrt{x} \cdot d x\right)\)
\(=\int(1 \cdot d x)-\left(\frac{4}{\pi}\right) \int\left(\cos ^{-1} \sqrt{x} \cdot d x\right)\)
\(\Rightarrow \mathrm{I}=x-\left(\frac{4}{\pi}\right) \mathrm{I}_{1}\quad \quad \ldots\text{(2)}\)
Now, first solve for \( \mathrm{I}_{1} \) :
as, \( \mathrm{I}_{1}=\int\left(\cos ^{-1} \sqrt{x} \cdot d x\right) \)
let \( \sqrt{x}-=t \)
\(\Rightarrow \frac{1}{2} x^{-\frac{1}{2}} d x=d t \)
\(\Rightarrow \frac{d x}{\sqrt{x}}=2 . d t \)
\(\Rightarrow d x=2 . t . d t \)
\(\Rightarrow \mathrm{I}_{1}=\int\left(\cos ^{-1} t .2 . t . d t\right)\)
\(=2 \int t \cdot \cos ^{-1} t d t\)
because, \( \int u \cdot v d x=u . \int v d x-\int \frac{d u}{d x} \cdot\left\{\int v d x\right\} d x \)
\(\Rightarrow \int t \cdot \cos ^{-1} t \cdot d t=2 \cdot\left[\cos ^{-1} t \cdot \int t \cdot d t-\right.\)\(\left.\int \frac{d\left(\cos ^{-1} t\right)}{d t} \cdot\left\{\int d t\right\} d t\right]\)
\(=2 \cos ^{-1} t \cdot \frac{t^{2}}{2}-2 \cdot \int\left(-\frac{1}{\sqrt{1-t^{2}}}\right) \cdot\left\{\frac{t^{2}}{2}\right\} d t\)
\(=t^{2} \cdot \cos ^{-1} t-\int\left(-\frac{1}{\sqrt{1-t^{2}}}\right) \cdot d t\)
\(=t^{2} \cdot \cos ^{-1} t-\int\left(\frac{-1+1-t^{2}}{\sqrt{1-t^{2}}}\right) \cdot d t\)
\(=t^{2} \cdot \cos ^{-1} t-\int\left(\frac{-1}{\sqrt{1-t^{2}}}+\frac{1-t^{2}}{\sqrt{1-t^{2}}}\right) \cdot d t\)
\(=t^{2} \cdot \cos ^{-1} t-\int\left(\frac{-1}{\sqrt{1-t^{2}}} d t\right)-\frac{t}{2} \cdot \sqrt{1-t^{2}}\)
\(\text { as, } \int\left(\sqrt{a^{2}-x^{2}}\right) \cdot d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)\)
\(\Rightarrow \mathrm{I}_{1}=t^{2} \cdot \cos ^{-1} t+\sin ^{-1} t-\frac{t}{2} \sqrt{1-t^{2}}-\frac{1}{2} \sin ^{-1}(t)\)
\(\Rightarrow \mathrm{I}_{1}=t^{2} \cdot \cos ^{-1} t-\frac{t}{2} \sqrt{1-t^{2}}-\frac{1}{2} \sin ^{-1} t\)
Put it in equ. (2)
\(\Rightarrow \mathrm{I}=x-\left(\frac{4}{\pi}\right)\left[t^{2} \cdot \cos ^{-1} t-\frac{t}{2} \sqrt{1-t^{2}}\right.\)\(\left.-\frac{1}{2} \sin ^{-1} t\right]\quad\ldots \text{(2)}\)
\(\Rightarrow \mathrm{I}=x-\left(\frac{4}{\pi}\right)\left[(\sqrt{x})^{2} \cdot \cos ^{-1} \sqrt{x}-\frac{\sqrt{x}}{2} \sqrt{1- (\sqrt{x})^{2}}\right.\)\(\left.-\frac{1}{2} \sin ^{-1} \sqrt{x}\right]\)
\(=x-\left(\frac{4}{\pi}\right)\left[x \cdot \cos ^{-1} \sqrt{x}-\frac{\sqrt{x}}{2} \sqrt{1-x}-\frac{1}{2} \sin ^{-1} \sqrt{x}\right]\)
\(=x-\left(\frac{4}{\pi}\right)\left[x \cdot\left(\frac{\pi}{2}-\sin ^{-1} \sqrt{x}\right)-\frac{\left(\sqrt{x-x^{2}}\right)}{2}\right.\)\(\left. \sqrt{1-x}-\frac{1}{2} \sin ^{-1} \sqrt{x}\right]\)
\(=x-2 x+\frac{4 x}{\pi} \sin ^{-1} \sqrt{x}+\frac{2}{\pi} \sqrt{x-x^{2}}-\frac{2}{\pi} \sin ^{-1} \sqrt{x}\)
\(=x+\frac{2}{\pi}\left[\frac{2(2 x-1)}{\pi} \sin ^{-1} \sqrt{x}\right]+\frac{2}{\pi} \sqrt{x-x^{2}}-x+C\)
\(\Rightarrow \mathrm{I}=x+\frac{2}{\pi}\left[\frac{2(2 x-1)}{\pi} \sin ^{-1} \sqrt{x}\right]+\frac{2}{\pi} \sqrt{x-x^{2}}-x+C\)

Given: \( \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \)
Let \( \mathrm{I}=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \)
Let \( x=\cos ^{2} \theta \Rightarrow \mathrm{d}x=-2 \sin \theta \cos \theta \mathrm{d} \theta \)
\(\Rightarrow \sqrt{x}=\cos \theta \text { or } \theta=\cos ^{-1} \sqrt{x}\)
\(\Rightarrow \mathrm{I}=\int \sqrt{\frac{1-\sqrt{\cos ^{2} \theta}}{1+\sqrt{\cos ^{2} \theta}}}(-2 \sin \theta \cos \theta) d \theta\)
\(=\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta\)
\(=\int-\sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}}(2 \sin \theta \cos \theta) d \theta\)
\(=\int-\sqrt{\frac{\sin ^{2} \frac{\theta}{2}}{\cos ^{2} \frac{\theta}{2}}}\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right) d \theta\)
\(=\int-\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} \cdot(2) \cdot\left(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right) \cdot\left(2 \cos \left(\frac{\theta}{2}\right)-1\right) d \theta\)
\(=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=\int-4 \cdot\left[\sin ^{2}\left(\frac{\theta}{2}\right)\right]\left(2 \cos \left(\frac{\theta}{2}\right)-1\right) d \theta\)
\(=\int-4 \cdot\left\{\left[2 \cdot \sin ^{2}\left(\frac{\theta}{2}\right) \cos ^{2}\left(\frac{\theta}{2}\right)\right]-\sin ^{2}\left(\frac{\theta}{2}\right)\right\} d \theta\)
\(=\int-2 \cdot\left(2 \cdot \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)^{2} d \theta+4 \int \sin ^{2}\left(\frac{\theta}{2}\right) \mathrm{d} \theta\)
\(=-2 \cdot \int \sin ^{2} \theta d \theta+4 \int \sin ^{2}\left(\frac{\theta}{2}\right) \mathrm{d} \theta\)
\(=-2 \cdot \int \frac{1-\cos 2 \theta}{2} d \theta+4 \int \frac{1-\cos \theta}{2} \mathrm{~d} \theta\)
\(=-2\left[\frac{\theta}{2}-\frac{\sin 2 \theta}{2}\right]+4\left[\frac{\theta}{2}-\frac{\sin \theta}{2}\right]+C\)
\(=-\theta+\frac{\sin 2 \theta}{2}+2 \theta-2 \sin \theta+C\)
\(=\theta+\frac{2 \cdot \sqrt{1-\cos ^{2} \theta} \cdot \cos \theta}{2}-2 \sqrt{1-\cos ^{2} \theta}+C\)
\(=\cos ^{-1} \sqrt{x}+\sqrt{1+x} \cdot \sqrt{x}-2 \sqrt{1-x}+C\)
\(=\cos ^{-1} \sqrt{x}+\sqrt{x(1+x)}-2 \sqrt{1-x}+C\)
\(\Rightarrow \mathrm{I}=\cos ^{-1} \sqrt{x}+\sqrt{x-x^{2}}-2 \sqrt{1-x}+C\)

Given: \( \frac{2+\sin 2 x}{1+\cos 2 x} e^{x} \)
Let \( \mathrm{I}=\frac{2+\sin 2 x}{1+\cos 2 x} e^{x} \)
\(=\left(\frac{2+2 \sin x \cos x}{2 \cos ^{2} x}\right) e^{x}\)
\(=2 \cdot\left(\frac{1+\sin x \cos x}{2 \cos ^{2} x}\right) e^{x}\)
\(=\left(\frac{1}{\cos ^{2} x}+\frac{\sin x \cos x}{\cos ^{2} x}\right) e^{x}\)
\(=\left(\sec ^{2} x+\tan x\right) e^{x}\)
Now let \( \tan x=f(x) \)
\(\Rightarrow \mathrm{f}^{\prime}(x)=\sec ^{2} \mathrm{d}x\)
\(\Rightarrow \int \frac{2+\sin 2 x}{1+\cos 2 x} e^{x} d x=\int\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x) e^{x} d x\right.\)
\(\Rightarrow e^{x} \mathrm{f}(x)+C\)
\(\Rightarrow \mathrm{I}=e^{x} \tan x+C\)

Given: \( \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} \)
Let \( \mathrm{I}=\frac{x^{2}+x+1}{(x+1)^{2}(x+2)} \)
Using partial differentiation:
Let \( \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)} \quad \quad \ldots\text{(1)}\)
\(\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A(x+1)(x+2)+B(x+2)+C(x+1)^{2}}{(x+1)^{2}(x+2)}\)
\(\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A\left(x^{2}+3 x+2\right)+B(x+2)+C\left(x^{2}+2 x+1\right)}{(x+1)^{2}(x+2)}\)
\(\Rightarrow x^{2}+x+1=\mathrm{A}x^{2}+3 \mathrm{A}x+2 \mathrm{A}+\mathrm{B}x+2 \mathrm{B}+\mathrm{C}x^{2}\)\(+2 \mathrm{C}x+\mathrm{C}\)
\(\Rightarrow x^{2}+x+1=(2 \mathrm{A}+2 \mathrm{B}+\mathrm{C})+(3 \mathrm{A}+\mathrm{B}+2 \mathrm{C}) x\)\(+(\mathrm{A}+\mathrm{C}) x^{2}\)
Equating the coefficients of \( x, x^{2} \) and constant value. We get:
(a) \( \mathrm{A}+\mathrm{C}=1 \)
(b) \( 3 \mathrm{A}+\mathrm{B}+2 \mathrm{C}=1 \)
(c) \( 2 \mathrm{A}+2 \mathrm{B}+\mathrm{C}=1 \)
After solving we get:
\(\mathrm{A}=-2, \mathrm{~B}=1 \text { and } \mathrm{C}=3\)
\(\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}\)
\(\Rightarrow \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x=\int\left(\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}\right)\)
\(=-2 \cdot \int\left(\frac{1}{x+1}\right) d x+\int\left(\frac{1}{(x+1)^{2}}\right) d x+3 \cdot \int\left(\frac{1}{(x+2)}\right) d x\)
\(=-2 \cdot \int\left(\frac{1}{x+1}\right) d x+\int\left((x+1)^{-2}\right) d x+3 \cdot \)\(\int\left(\frac{1}{(x+2)}\right) d x\)
\(=-2 \log |x+1|+\left(\frac{(x+1)^{-1}}{(-1)}\right)+3 \log |x+1|+C\)
\(=-2 \log |x+1|-\frac{1}{(x+1)}+3 \log |x+1|+C\)

Given: \( \tan ^{-1} \sqrt{\frac{1-x}{1+x}} \)
Let \( \mathrm{I}=\tan ^{-1} \sqrt{\frac{1-x}{1+x}} \)
Let \( x=\cos \theta \Rightarrow d x=-\sin \theta d \theta \)
\(\Rightarrow \theta=\cos ^{-1} x\)
\(\Rightarrow \mathrm{I}=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} \int \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta) d \theta\)
\(=\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)}{2 \cos ^{2}\left(\frac{\theta}{2}\right)}}(\sin \theta) d \theta\)
\(=\int \tan ^{-1} \sqrt{\tan \left(\frac{\theta}{2}\right)}(\sin \theta) d \theta\)
\(=\int \tan ^{-1} \tan \frac{\theta}{2} \cdot(\sin \theta) d \theta\)
\(=-\frac{1}{2} \int \theta \cdot(\sin \theta) d \theta\)
Because, \( \int u . v d x=u . \int v d x-\int \frac{d u}{d x} \cdot\left\{\int v d x\right\} d x \)
\(=-\frac{1}{2} \int \theta \cdot(\sin \theta) d \theta=-\frac{1}{2}\left[\theta \cdot \int \sin \theta d \theta-\int \frac{d \theta}{d \theta} \cdot\left\{\int \sin v d \theta\right\} d \theta\right]\)
\(=-\frac{1}{2}\left[\theta \cdot(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta\right]\)
\(=-\frac{1}{2}[-\theta \cdot \cos \theta+\sin \theta]\)
\(=\frac{1}{2} \theta \cdot \cos \theta-\frac{1}{2} \sqrt{\left(1-\cos ^{2} \theta\right)}\)
\(=\frac{1}{2} \cos ^{-1} x \cdot x-\frac{1}{2} \sqrt{\left(1-x^{2}\right)}+C\)

Given: \( \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}} \)
Let \( \mathrm{I}=\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}} \)
\(=\frac{\sqrt{x^{2}+1}}{x^{4}}\left[\log \left(x^{2}+1\right)-\log x^{2}\right]\)
Now let \( 1+\frac{1}{x^{2}}=t \Rightarrow-\frac{2}{x^{3}} d x=d t \)
\(\Rightarrow \int \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}} d x=\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}}\left[\log \left(x+\frac{1}{x^{2}}\right)\right] d x\)
\(=\int-\frac{1}{2} \cdot \sqrt{t}[\log (t)] d t\)
Because, \( \int u . v d x=u . \int v d x-\int \frac{d u}{d x} \cdot\left\{\int v d x\right\} d x \)
\(=\int-\frac{1}{2} \cdot \sqrt{t}[\log (t)] d t=-\frac{1}{2}\left[\log t . \int \sqrt{t} d t-\int \frac{d}{d t} \log t \cdot\left\{\int \sqrt{t} d t\right\} d t\right]\)
\(=-\frac{1}{2}\left[\log t \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{t} \cdot\left\{\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right\} d t\right]\)
\(=-\frac{1}{2}\left[\frac{2}{3} t^{\frac{3}{2}} \log t-\int\left\{\frac{t^{\frac{3}{2}}-1}{\frac{3}{2}}\right\} d t\right]\)
\(=-\frac{1}{2}\left[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{2}{3} \int t^{\frac{1}{2}} d t\right]\)
\(=-\frac{1}{2}\left[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{2}{3} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]\)
\(=\left[-\frac{1}{2} \cdot \frac{2}{3} t^{\frac{3}{2}} \log t+\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot t^{\frac{3}{2}}\right]\)
\(=-\frac{1}{3} t^{\frac{3}{2}}\left[\log t-\frac{2}{3}\right]\)
\(\Rightarrow \mathrm{I}=-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}\left[\log \left(1+\frac{1}{x^{2}}\right)-\frac{2}{3}\right]+C\)

Given: \( \int_{-\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x \)
Let \( \mathrm{I}=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x \)
\( =\int_{-\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2}\left(\frac{x}{2}\right)}\right) d x \)
\( =\int_{-\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1}{2 \sin ^{2}\left(\frac{x}{2}\right)}-\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2}\left(\frac{x}{2}\right)}\right) d x \)
\( =\int_{-\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1}{2} \operatorname{cosec}^{2}\left(\frac{x}{2}\right)-\cot \frac{x}{2}\right) d x \)
Now let \( \mathrm{f}(x)=-\cot \frac{x}{2} \)
\(\Rightarrow \mathrm{f}^{\prime}(x)=-\left(\frac{1}{2} \operatorname{cosec}^{2}\left(\frac{x}{2}\right)\right)=\frac{1}{2} \operatorname{cosec}^{2}\left(\frac{x}{2}\right)\)
\(\Rightarrow \int_{-\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1}{2} \operatorname{cosec}^{2}\left(\frac{x}{2}\right)-\cot \frac{x}{2}\right) d x=\int_{-\frac{\pi}{2}}^{\pi}\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right) e^{x} d x\)
\(=\left[e^{x} \mathrm{f}(x)\right]_{-\frac{\pi}{2}}^{\pi}\)
\(=-\left[e^{\pi}\left(\cot \frac{\pi}{2}\right)-e^{\frac{\pi}{2}}\left(\cot \frac{\pi}{4}\right)\right]\)
\(=-\left[e^{\pi}(0)-e^{\frac{\pi}{2}}(1)\right]\)
\(=-\left[0-e^{\frac{\pi}{2}}\right]\)
\(\Rightarrow \mathrm{I}=e^{\frac{\pi}{2}}\)

Given: \( \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x \)
\( =\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\left(1+\frac{\sin ^{4} x}{\cos ^{4} x}\right)} d x \)
\( =\int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec ^{2} x}{\left(1+\tan ^{4} x\right)} d x \)
Now let \( \tan ^{2}x =\mathrm{t} \Rightarrow 2 \tan x \sec ^{2}x d x=\mathrm{dt} \)
And when \( x=0 \) then \( \mathrm{t}=0 \) and when \(x =\frac{ \pi }{ 4 } \) then \( \mathrm{t}=1 \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec ^{2} x}{\left(1+\tan ^{4} x\right)} d x=\int_{0}^{1} \frac{1}{\left(1+t^{2}\right)}\left(\frac{d t}{2}\right)\)
\(\Rightarrow \mathrm{I}=\frac{1}{2}\left[\tan ^{-1} t\right]_{0}^{1}\)
\(\Rightarrow \frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]\)
\(\Rightarrow \mathrm{I}=\frac{1}{2} \cdot \frac{\pi}{4}\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{8}\)

Given: \( \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x \quad\ldots \text{(1)}\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4\left(1-\cos ^{2} x\right)} d x\)
\(=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4(1)-\left(4 \cos ^{2} x\right)} d x\)
\(=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{4-3 \cos ^{2} x} d x\)
\(=\int_{0}^{\frac{\pi}{2}} \frac{\frac{1}{3} \cdot 3 \cos ^{2} x}{4-3 \cos ^{2} x} d x\)
\(=-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{-3 \cos ^{2} x+4-4}{4-3 \cos ^{2} x} d x\)
\(=-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x-4}{4-3 \cos ^{2} x} d x\)
\(=-\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x}{4-3 \cos ^{2} x} d x+\frac{1}{3} \int_{0}^{\frac{\pi}{2}} \frac{4}{4-3 \cos ^{2} x} d x\)
\(=-\frac{1}{3} \cdot\left[\frac{\pi}{2}\right]+\frac{1}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{4 \sec ^{2} x}{4\left(1+\tan ^{2} x\right)-3} d x\)
\(\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x\)
\(\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\mathrm{I}_{1}\quad\ldots \text{(2)}\)
First solve for \( \mathrm{I}_{1} \) :
\(\mathrm{I}_{1}=\frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x\)
Let \( 2 \tan x =\mathrm{t} \Rightarrow 2 \sec ^{2} x\mathrm{d}x\mathrm{dt} \)
When \( x=0 \) then \( \mathrm{t}=0 \) and when \( x=\frac{ \pi }{ 2 } \) then \( \mathrm{t}=\infty \)
\(\Rightarrow \frac{2}{3} \cdot \int_{0}^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x=\frac{2}{3} \cdot \int_{0}^{\infty} \frac{1}{1+t^{2}} d t\)
\(\Rightarrow \mathrm{I}_{1}=\frac{2}{3}\left[\tan ^{-1} t\right]_{0}^{\infty}\)
\(=\frac{2}{3}\left[\tan ^{-1} \infty \tan ^{-1} 0\right]\)
\(\Rightarrow \mathrm{I}_{1}=\frac{2}{3} \cdot \frac{\pi}{2}\)
\(\Rightarrow \mathrm{I}_{1}=\frac{\pi}{3}\)
Put this value in equ. (2)
\(\Rightarrow \mathrm{I}=-\frac{\pi}{6}+\frac{\pi}{3}\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{6}\)

Given: \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x \)
Let \( \mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x \)
\( =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{-(-\sin 2 x)}} d x \)
\( =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{-(-1+1-2 \sin x \cos x)}} d x \)
\( =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)}} d x \)
\( =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\left(1-(\sin x-\cos x)^{2}\right)}} d x \)
Now let \( \sin x-\cos x =\mathrm{t} \Rightarrow(\cos x+\sin x ) \mathrm{d}x=\mathrm{dt} \)
When \( x=\frac{\pi}{6} \)
\(\Rightarrow t=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2} \) and
when \( x=\frac{\pi}{3} \)
\(\Rightarrow t=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2} \)
\( =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\left(1-(\sin x-\cos x)^{2}\right)}} d x=\int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{1}{\sqrt{\left(1-(t)^{2}\right)}} d t \)
\(=\int_{-\left(\frac{\sqrt{3}-1}{2}\right)}^{\frac{\sqrt{3}-1}{2}} \frac{1}{\sqrt{\left(1-(t)^{2}\right)}} d t\)
Let \( \mathrm{f}(x)=\frac{1}{\sqrt{\left(1-(t)^{2}\right)}} \)
and \( \mathrm{f}(-x)=\frac{1}{\sqrt{\left(1-(t)^{2}\right)}}=\frac{1}{\sqrt{\left(1-(t)^{2}\right)}}=\mathrm{f}(x) \)
i.e. \( f(x)=f(-x) \)
so, \( f(x) \) is an even function.
It is also known that if \( f(x) \) is an even function then,
\(\left\{\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\right\}\)
\(\Rightarrow \mathrm{I}=2 \cdot \int_{0}^{\frac{\sqrt{3}-1}{2}} \frac{1}{\sqrt{\left(1-(t)^{2}\right)}} d t\)
\(\Rightarrow \mathrm{I}=\left[2 \cdot \sin ^{-1} t\right]_{0}^{\frac{\sqrt{3}-1}{2}}\)
\(\Rightarrow \mathrm{I}=2 \cdot \sin ^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\)

Given: \( \int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}} \)
Let \( \mathrm{I}=\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}} \)
\(=\int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} \times \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}} d x\)
\(=\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} d x\)
\(=\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1} d x\)
\(=\int_{0}^{1} \sqrt{1+x} d x+\int_{0}^{1} \sqrt{x} d x\)
\(=\int_{0}^{1}\left((1+x)^{\frac{1}{2}}\right) d x+\int_{0}^{1}(x)^{\frac{1}{2}} d x\)
\(\Rightarrow \mathrm{I}=\left[\frac{(1+x)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}+\left[\frac{(x)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\)
\(=\frac{2}{3} \cdot\left[(1+1)^{\frac{3}{2}}-(1+0)^{\frac{3}{2}}\right]+\frac{2}{3} \cdot\left[(1)^{\frac{3}{2}}\right]\)
\(=\frac{2}{3} \cdot\left[(2)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]+\frac{2}{3} \cdot\left[(1)^{\frac{3}{2}}\right]\)
\(=\frac{2}{3} \cdot\left[(2)^{\frac{3}{2}}-1\right]+\frac{2}{3} \cdot[1]\)
\(=\frac{2}{3} \cdot\left[(2)^{\frac{3}{2}}\right]-\frac{2}{3} \cdot[1]+\frac{2}{3} \cdot[1]\)
\(=\frac{2}{3} \cdot[2 \sqrt{2}]\)
\(=\mathrm{I}=\frac{4 \sqrt{2}}{3}\)

Given: \( \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \)
Also, let \( \sin -\cos =\mathrm{t} \)
Differentiating both sides, we get,
\((\cos x+\sin x) d x=d t\)
When \( x=0, t=-1 \)
And when \( x=\frac{\pi}{4}, \mathrm{t}=0 \)
Now,
\((\sin x-\cos x)^{2}=t^{2}\)
\(1-2 \sin x \cdot \cos x=t^{2}\)
\( \operatorname{Sin} 2 x=1-t^{2} \)
Putting all the values, we get the integral,
\(I=\int_{-1}^{0} \frac{d t}{9+16\left(1-t^{2}\right)}\)
\(I=\int_{-1}^{0} \frac{d t}{25-16 t^{2}}\)
\(I=\int_{-1}^{0} \frac{d t}{(5)^{2}-(4 t)^{2}}\)
\(I=\frac{1}{4}\left[\frac{1}{2(5)} \log \left|\frac{5+4 t}{5-4 t}\right|\right]_{-1}^{0}\)
\(I=\frac{1}{40}\left[\log 1-\log \frac{1}{9}\right]\)
\(I=\frac{1}{40} \log 9\)

Given: \( \int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x \)
\(=\int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \cdot \tan ^{-1}(\sin x) d x\)
Let \( \sin =\mathrm{t} \Rightarrow \cos x\mathrm{d}x=\mathrm{dt} \)
When \( x=0 \) then \( \mathrm{t}=0 \) and when \( x=\frac{ \pi }{ 2 } \) then \( \mathrm{t}=1 \)
\(\Rightarrow \int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \cdot \tan ^{-1}(\sin x) d x=\int_{0}^{1} 2 t \cdot \tan ^{-1}(t) d t\)
Because, \( \int u . v d x=u . \int v d x-\int \frac{d u}{d x} \cdot\left\{\int v d x\right\} d x \)
\(\Rightarrow 2 \int_{0}^{1} t \cdot \tan ^{-1}(t) d t=2\left[\tan ^{-1}(t) \cdot \int \mathrm{tdt}-\int \frac{\mathrm{d}}{\mathrm{dt}}\left(\tan ^{-1}(t)\right) \right.\)\(\left.\cdot\left\{\int \mathrm{t} \cdot \mathrm{dt}\right\} \mathrm{dt}\right]\)
\(=2\left[\tan ^{-1}(t) \cdot \frac{t^{2}}{2}-\int \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t\right]\)
\(=2\left[\tan ^{-1}(t) \cdot \frac{t^{2}}{2}-\frac{1}{2} \cdot \int \frac{-1+1+t^{2}}{1+t^{2}} d t\right]\)
\(=2\left[\tan ^{-1}(t) \cdot \frac{t^{2}}{2}-\frac{1}{2} \cdot\left\{\int-\frac{1}{1+t^{2}} d t+\int \frac{1+t^{2}}{1+t^{2}} d t\right\}\right]\)
\(=2\left[\tan ^{-1}(t) \cdot \frac{t^{2}}{2}-\frac{1}{2} \cdot\left\{\int-\frac{1}{1+t^{2}} d t+\int 1 d t\right\}\right]\)
\(=2\left[\tan ^{-1}(t) \cdot \frac{t^{2}}{2}-\frac{1}{2} \cdot\left\{-\tan ^{-1}(t)+t\right\}\right]\)
\(=\left[t^{2} \cdot \tan ^{-1}(t) \cdot-\left\{-\tan ^{-1}(t)+t\right\}\right]\)
\(\Rightarrow 2 \int_{0}^{1} \tan ^{-1}(t) d t=\left[t^{2} \cdot \tan ^{-1}(t) \cdot-\left\{-\tan ^{-1}(t)+t\right\}\right]_{0}^{1}\)
\(=\left[1^{2} \cdot \tan ^{-1}(1) \cdot-\left\{-\tan ^{-1}(1)+1\right\}\right]-\left[0^{2} \cdot \tan ^{-1}(0)+0\right]\)
\(=\left[1 \cdot \frac{\pi}{4} \cdot-\left\{-\frac{\pi}{4}+1\right\}\right]\)
\(=\left[\frac{\pi}{4}+\frac{\pi}{4}-1\right]\)
\(\Rightarrow \mathrm{I}=\left[\frac{\pi}{2}-1\right]\)

Given: \( \int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x \)
Let \( \mathrm{I}=\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x\quad\ldots \text{(1)} \)
as, \( \left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\pi} \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} d x\)
\(=\int_{0}^{\pi} \frac{(\pi-x)(-\tan x)}{(-\sec x)+(-\tan x)} d x\)
\(=\int_{0}^{\pi} \frac{-(\pi-x)(\tan x)}{-[(-\sec x)+(-\tan x)]} d x\)
\(=\int_{0}^{\pi} \frac{(\pi-x)(-\tan x)}{\sec x+\tan x} d x\quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x}+\frac{(\pi-x)(-\tan x)}{\sec x+\tan x} d x\)
\(2 \mathrm{I}=\int_{0}^{\pi} \frac{\pi \tan x}{\sec x+\tan x} d x\)
\(=\int_{0}^{\pi} \frac{\pi \frac{\sin x}{\cos x}}{\frac{1}{\cos x} \frac{\sin x}{\cos x}} d x\)
\(2 \mathrm{I}=\pi \cdot \int_{0}^{\pi} \frac{(\sin x)}{(1+\sin x)} d x\)
\(=\pi \cdot \int_{0}^{\pi} \frac{(-1+1+\sin x)}{(1+\sin x)} d x\)
\(=\pi \cdot \int_{0}^{\pi} \frac{(-1)}{(1+\sin x)} d x+\pi \cdot \int_{0}^{\pi} \frac{(1+\sin x)}{(1+\sin x)} d x\)
\(=\pi \cdot \int_{0}^{\pi} \frac{(-1)}{(1+\sin x)} \times \frac{(1-\sin x)}{(1-\sin x)} d x+\pi \cdot \int_{0}^{\pi} 1 \cdot d x\)
\(=-\pi \cdot \int_{0}^{\pi} \frac{(1+\sin x)}{(1+\sin 2 x)} d x+\pi \cdot \int_{0}^{\pi} 1 \cdot d x\)
\(2 \mathrm{I}=-\pi \cdot \int_{0}^{\pi} \frac{(1-\sin x)}{\cos ^{2} x} d x+\pi \cdot \int_{0}^{\pi} 1 \cdot d x\)
\(2 \mathrm{I}=-\pi \cdot \int_{0}^{\pi}\left\{\frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}\right\} d x+\pi \cdot \int_{0}^{\pi} 1 \cdot d x\)
\(2 \mathrm{I}=-\pi \cdot \int_{0}^{\pi}\left\{\sec ^{2} x-\tan x \sec x\right\} d x+\pi \cdot \int_{0}^{\pi} 1 \cdot d x\)
\(\Rightarrow 2 \mathrm{I}=-\pi \cdot[\tan x-\sec x]_{0}^{\pi}+[x]_{0}^{\pi}\)
\(\Rightarrow 2 \mathrm{I}=-\pi \cdot[0-(-1)-0+1]+\pi \cdot[\pi]\)
\(\Rightarrow 2 \mathrm{I}=\pi \cdot[-2+\pi]\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{2} \cdot[\pi-2]\)

Given: \( \int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x \)
\(\Rightarrow \mathrm{I}=\int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x\)
\(\Rightarrow \mathrm{I}=\int_{1}^{4}[|x-1|] d x+\int_{1}^{4}[|x-2|] d x+\int_{1}^{4}[|x-3|] d x\)
Let \( \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3} \)
First solve for I1:
\(\mathrm{I}_{1}=\int_{1}^{4}[|x-1|] d x\)
As we can see that \( (-1) \geq 0 \) when \( 1 \leq x \leq 4 \)
\(\Rightarrow \mathrm{I}_{1}=\int_{1}^{4}(x-1) d x\)
\(\Rightarrow \mathrm{I}_{1}=\left[\frac{x^{2}}{2}-x\right]_{0}^{1}\)
\(\Rightarrow \mathrm{I}_{1}=\left[\frac{(4)^{2}}{2}-4-\frac{(1)^{2}}{2}-1\right]\)
\(\Rightarrow \mathrm{I}_{1}=\left[8-4-\frac{1}{2}+1\right]\)
\(\Rightarrow \mathrm{I}_{1}=\left[5-\frac{1}{2}\right]\)
\(\Rightarrow \mathrm{I}_{1}=\frac{9}{2}\)
Now solve for \( \mathrm{I}_{2} \) :
\(\mathrm{I}_{2}=\int_{1}^{4}[|x-2|] d x\)
As we can see that \( (-2) \leq 0 \) when \( 1 \leq x \leq 2 \) and \( (-2) \geq 0 \) when \( 2 \leq x \leq 4 \)
\(\text {as, }\left\{\int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\int_{a}^{c} \mathrm{f}(x) \mathrm{d}x+\int_{c}^{b} \mathrm{f}(x) \mathrm{d}x\right\}\)
\(\Rightarrow \mathrm{I}_{2}=\int_{1}^{2}-(x-2) d x+\int_{2}^{4}(x-2) d x\)
\(\Rightarrow \mathrm{I}_{2}=\left[\frac{x^{2}}{2}-2 x\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2 x\right]_{2}^{4}\)
\(\Rightarrow \mathrm{I}_{2}=\left[\frac{(2)^{2}}{2}-2(2)-\frac{(1)^{2}}{2}-2(1)\right]\)
\(\Rightarrow \mathrm{I}_{2}=\left[2-4-\frac{1}{2}+2\right]+[8-8-2+4]\)
\(\Rightarrow \mathrm{I}_{2}=\left[\frac{1}{2}+2\right]\)
\(\Rightarrow \mathrm{I}_{2}=\frac{5}{2}\)
Now solve for \( \mathrm{I}_{3} \) :
\(\mathrm{I}_{3}=\int_{1}^{4}[|x-3|] d x\)
As we can see that \( (x-3) \leq 0 \) when \( 1 \leq x \leq 3 \) and \( (x-3) \geq 0 \) when \( 3 \leq x \leq 4 \)
as, \( \left\{\int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=\int_{a}^{c} \mathrm{f}(x) \mathrm{d}x+\int_{c}^{b} \mathrm{f}(x) \mathrm{d}x\right\} \)
\(\Rightarrow \mathrm{I}_{3}=\int_{1}^{3}-(x-3) d x+\int_{3}^{4}(x-3) d x\)
\(\Rightarrow \mathrm{I}_{3}=\left[\frac{x^{2}}{2}-3 x\right]_{1}^{3}+\left[\frac{x^{2}}{2}-3 x\right]_{3}^{4}\)
\(\Rightarrow \mathrm{I}_{3}=\left[\frac{(3)^{2}}{2}-3(3)-\frac{(1)^{2}}{2}-3(1)\right]\)
\(\Rightarrow \mathrm{I}_{3}=\left[\frac{9}{2}-9-\frac{1}{2}+3\right]+\left[8-12-\frac{9}{2}+9\right]\)
\(\Rightarrow \mathrm{I}_{3}=\left[2+\frac{1}{2}\right]\)
\(\Rightarrow \mathrm{I}_{3}=\frac{5}{2}\)
\(\text { as } \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}\)
\(\Rightarrow \mathrm{I}=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}\)
\(\Rightarrow \mathrm{I}=\frac{19}{2}\)

Given: \( \int_{1}^{3} \frac{d x}{\left(x^{2}\right)(x+1)} \)
To Prove: \( \int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3} \)
Let \( \mathrm{I}=\frac{d x}{\left(x^{2}\right)(x+1)} \)
Using partial differentiation:
Let \( \frac{1}{\left(x^{2}\right)(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1} \quad\ldots \text{(1)}\)
\(\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{A(x)(x+1)+B(x+1)+C\left(x^{2}\right)}{(x+1)\left(x^{2}\right)}\)
\(\Rightarrow 1=A\left(x^{2}+x\right)+(B x+B)+C x^{2}\)
\(\Rightarrow 1=\mathrm{A} x^{2}+\mathrm{A}x+\mathrm{B}+\mathrm{B}x+\mathrm{C}x^{2}\)
\(\Rightarrow 1=\mathrm{B}+(\mathrm{A}+\mathrm{B}) x+(\mathrm{A}+\mathrm{C}) x^{2}\)
Equating the coefficients of \( x, x^{2} \) and constant value. We get:
(a) \( \mathrm{B}=1 \)
(b) \( \mathrm{A}+\mathrm{B}=0 \Rightarrow \mathrm{A}=-\mathrm{B} \Rightarrow \mathrm{A}=-1 \)
(c) \( \mathrm{A}+\mathrm{C}=0 \Rightarrow \mathrm{C}=-\mathrm{A} \Rightarrow \mathrm{C}=1 \)
Put these values in equation (1)
\(\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}\)
\(\Rightarrow \frac{1}{\left(x^{2}\right)(x+1)}=\frac{-1}{x}+\frac{1}{x^{2}}+\frac{1}{x+1}\)
\(\Rightarrow \int \frac{1}{\left(x^{2}\right)(x+1)} d x=\int-\frac{1}{x} d x+\int \frac{1}{\left(x^{2}\right)} d x+\int \frac{1}{(x+1)} d x\)
\(\Rightarrow \int_{1}^{3} \frac{1}{\left(x^{2}\right)(x+1)} d x=\left[-\log |x|-x^{-1}+\log |x+1|\right]_{1}^{3}\)
\(\Rightarrow \int_{1}^{3} \frac{1}{\left(x^{2}\right)(x+1)} d x=\left[-\frac{1}{x}+\log \left|\frac{x+1}{x}\right|\right]_{1}^{3}\)
\(=\left[-\frac{1}{3}+\log \left|\frac{3+1}{3}\right|-\left(-\frac{1}{1}+\log \left|\frac{1+1}{1}\right|\right)\right]\)
\(=\left[-\frac{1}{3}+\log \left|\frac{4}{3}\right|-\left(-\frac{1}{1}+\log \left|\frac{2}{1}\right|\right)\right]\)
\(=\left[-\frac{1}{3}+1+\log \left|\frac{4}{3} \times \frac{1}{2}\right|\right]\)
\(\Rightarrow \text { I }=\left[\frac{2}{3}+\log \left|\frac{2}{3}\right|\right]\)
\(\Rightarrow \text { L.H.S = R.H.S }\)
Hence proved.

Given: \( \int_{0}^{1} x e^{x} d x \)
To Prove: \( \int_{0}^{1} x e^{x} d x=1 \)
Let \( \mathrm{I}=\int_{0}^{1} x e^{x} d x \)
Because, \( \int u . v d x=u . \int v d x-\int \frac{d u}{d x} \cdot\left\{\int v d x\right\} d x \)
\(\Rightarrow \int_{0}^{1} x e^{x} d x= x\cdot \int_{0}^{1} e^{x} d x-\int_{0}^{1} \frac{d x}{d x} \cdot\left\{\int e^{x} \mathrm{d}x\right\} . \mathrm{d}x\)
\(\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 \cdot e^{x} d x\)
\(\Rightarrow \int_{0}^{1} x e^{x} d x=\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1}\)
\(\Rightarrow \int_{0}^{1} x e^{x} d x=\left[1 \cdot e^{1}-0 \cdot e^{0}\right]-\left[e^{1}-e^{0}\right]\)
\(\Rightarrow \int_{0}^{1} x e^{x} d x=e-0-e+1\)
\(\Rightarrow \int_{0}^{1} x e^{x} d x=1\)
L.H.S = R.H.S
Hence Proved.

Given: \( \int_{-1}^{1} x^{17} \cos ^{4} x d x \)
To Prove: \( \int_{-1}^{1} x^{17} \cos ^{4} x d x=0 \)
Let \( \mathrm{I}=\int_{-1}^{1} x^{17} \cos ^{4} x d x \)
As we can see \( f(x)=x^{17} \cdot \cos ^{4} x \)
and \( f(-x)=(-x)^{17} \cdot \cos ^{4}(-x)=-x^{17} \cdot \cos ^{4} x \)
i.e. \( f(x)=-f(-x) \)
so, it is an odd function.
It is also known that if \( f(x) \) is an odd function then,
\(\left\{\int_{-a}^{a} \mathrm{f}(x) \mathrm{d}x=0\right\}\)
\(\Rightarrow \mathrm{I}=\int_{-1}^{1} x^{17} \cos ^{4} x d x=0\)
Hence proved.

Given: \( \int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x \)
To Prove: \( \int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3} \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x \quad \quad \ldots\text{(1)}\)
\(=\int_{0}^{\frac{\pi}{2}} \sin x \cdot \sin ^{2} x d x\)
\(=\int_{0}^{\frac{\pi}{2}} \sin x \cdot\left(1-\cos ^{2} x\right) d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sin x d x-\int_{0}^{\frac{\pi}{2}} \sin x \cdot \cos ^{2} x d x\)
\(\Rightarrow \mathrm{I}=[-\cos x]_{0}^{\frac{\pi}{2}}-\mathrm{I}_{1}\quad \quad \ldots\text{(2)}\)
First solve for \( \mathrm{I}_{1} \) :
\(\Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \sin x \cdot \cos ^{2} x d x\)
Let \( \cos =\mathrm{t} \Rightarrow-\sin x\mathrm{d}x=\mathrm{dt} \Rightarrow \sin x\mathrm{d}x=-\mathrm{dt} \)
When \( x=0 \) then \( \mathrm{t}=1 \) and when \( x=\frac{ \pi }{ 2 } \) then \( \mathrm{t}=0 \)
\(\Rightarrow \mathrm{I}_{1}=\int_{1}^{0} t^{2}(-d t)\)
\(=-\int_{1}^{0} t^{2}(d t)\)
\(=-\left[\frac{t^{3}}{3}\right]_{1}^{0}\)
\(=-\left\{-\frac{1}{3}\right\}\)
\(\Rightarrow \mathrm{I}_{1}=\frac{1}{3}\)
Put in equ. (2)
\(\Rightarrow \mathrm{I}=[-\cos x]_{0}^{\frac{\pi}{2}}-\frac{1}{3}\)
\(\Rightarrow \mathrm{I}=-\left\{\cos \frac{\pi}{2}-\cos 0\right\}-\frac{1}{3}\)
\(\Rightarrow \mathrm{I}=1-\frac{1}{3}\)
\(\Rightarrow \mathrm{I}=\frac{2}{3}\)
L.H.S = R.H.S
Hence Proved.

Given: \( \int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x \)
To Prove: \( \int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2 \)
Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x \quad \quad \ldots\text{(1)} \)
\( =\int_{0}^{\frac{\pi}{4}} 2 \cdot \tan x \tan ^{2} x d x \)
\( =2 \int_{0}^{\frac{\pi}{4}} \tan x \cdot\left(\sec ^{2} x-1\right) d x \)
\( \Rightarrow \mathrm{I}=2\left\{-\int_{0}^{\frac{\pi}{4}} \tan x d x+\int_{0}^{\frac{\pi}{4}} \tan x \cdot \sec ^{2} x d x\right\} \)
\( \Rightarrow \mathrm{I}=[2 \log \sec x]_{0}^{\frac{\pi}{4}}+2 . \mathrm{I}_{1} \quad \quad \ldots\text{(2)}\)
First solve for \( \mathrm{I}_{1} \) :
\(\Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{4}} \tan x \cdot \sec ^{2} x d x\)
Let \( \tan x=\mathrm{t} \Rightarrow \sec ^{2} x\mathrm{d}x=\mathrm{dt} \)
When \( x=0 \) then \( \mathrm{t}=0 \) and when \(x =\frac{ \pi }{ 2 } \) then \( \mathrm{t}=1 \)
\(\Rightarrow \mathrm{I}_{1}=\int_{0}^{1} t . d t\)
\(=\left[\frac{t^{2}}{2}\right]_{0}^{1}\)
\(\Rightarrow \mathrm{I}_{1}=\frac{1}{2}\)
Put in equ. (2)
\(\Rightarrow \mathrm{I}=[2 \log \cos x]_{0}^{\frac{\pi}{4}}+2 \cdot \frac{1}{2}\)
\(\Rightarrow \mathrm{I}=2\left\{\log \cos \frac{\pi}{4}-\log \cos 0\right\}+1\)
\(\Rightarrow \mathrm{I}=2\left\{\log \frac{1}{\sqrt{2}}-\log 1\right\}+1\)
\(\Rightarrow \mathrm{I}=2\left\{\log \left(\frac{1}{\sqrt{2}}\right)^{2}-\log (1)^{2}\right\}+1\)
\(\Rightarrow \mathrm{I}=1-\log 2+\log 1\)
\(\Rightarrow \mathrm{I}=1-\log 2\)
\(\text { L.H.S }=\text { R.H.S }\)
Hence Proved.

Given: \( \int_{0}^{1} \sin ^{-1} x d x \)
To Prove: \( \int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1 \)
Let \( \mathrm{I}=\int_{0}^{1} \sin ^{-1} x d x \)
Because, \( \int u \cdot v d x=u . \int v d x-\int \frac{d u}{d x} \cdot\left\{\int v d x\right\} d x \)
\(\Rightarrow \int_{0}^{1} x e^{x} d x=\sin ^{-1} x \cdot \int_{0}^{1} 1 \cdot d x-\int_{0}^{1} \frac{d}{d x} \sin ^{-1} x\left\{\int 1 . \mathrm{d}x\right\} . \mathrm{d}x\)
\(\Rightarrow \int_{0}^{1} x e^{x} d x=\left[\sin ^{-1} x \cdot x\right]_{0}^{1}-\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} \cdot x d x\)
\(\Rightarrow \mathrm{I}=\left[\sin ^{-1} x \cdot x\right]_{0}^{1}-\mathrm{I}_{1}\quad\ldots \text{(2)}\)
First solve for \(\mathrm{I}_{1} \) :
\(\Rightarrow \mathrm{I}_{1}=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} \cdot x d x\)
Let \( 1-x^{2}=\mathrm{t} \Rightarrow-2 x\mathrm{d}x=\mathrm{dt} \)
When \( x=0 \) then \( \mathrm{t}=1 \) and when \(x =1 \) then \( \mathrm{t}=0 \)
\(\Rightarrow \mathrm{I}_{1}=\int_{0}^{1} \frac{1}{\sqrt{t}} \cdot \frac{-d t}{2}\)
\(=-\frac{1}{2}\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right]_{1}^{0}\)
\(\Rightarrow \mathrm{I}_{1}=\sqrt{1}\)
\(\Rightarrow \mathrm{I}_{1}=1\)
Put in equ. (2)
\(\Rightarrow \mathrm{I}=\left[\sin ^{-1} x \cdot x\right]_{0}^{1}-1\)
\(\Rightarrow \mathrm{I}=\sin ^{-1}(1)-0-1\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{2}-1\)
\(\text {L.H.S }=\text { R.H.S }\)
Hence Proved.

Given: \( \int_{0}^{1} e^{2-3 x} d x \)
Let \( \mathrm{I}=\int_{0}^{1} e^{2-3 x} d x \)
Because,
\( \int_{a}^{b} \mathrm{f}(x) \mathrm{d}x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[\mathrm{f}(\mathrm{a})+\mathrm{f} (\mathrm{a}+\mathrm{h})+\cdots+\mathrm{f}(\mathrm{a}\)\(+(\mathrm{n}-1)) h] \)
Where, \( \mathrm{h}=\frac{b-a}{n} \)
Here, \( a=0, b=1 \), and \( f(x)=e^{2-3 x} \) and \( h=\frac{1 }{ n} \)
\(\Rightarrow \int_{0}^{1} e^{2-3 x} d x\)
\(=(1-0) \lim _{n \rightarrow \infty} \frac{1}{n}[f(0)+f(0+h)+f(0+2 h)\)\(+\cdots+f(0+(n-1) h)]\)
\(=(1) \lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}+e^{2-3 h}+e^{2-3(2 h)}+\cdots+e^{2-3(n-1) h}\right]\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}+e^{2} \cdot e^{3 h}+e^{2} \cdot e^{-6 h} \ldots+e^{2} \cdot e^{-3(n-1) h}\right]\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}\left\{1+e^{3 h}+e^{-6 h} \ldots+e^{-3(n-1) h}\right\}\right]\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}\left\{\frac{1-\left(e^{-3 h}\right)^{n}}{1-\left(e^{-3 h}\right)}\right\}\right]\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}\left\{\frac{1-\left(e^{-\frac{3}{n}}\right)^{n}}{1-\left(e^{-\frac{3}{n}}\right)}\right\}\right] \mathrm{a}x, \mathrm{h}=\frac{1}{n}\)
\(=\lim _{n \rightarrow \infty} \frac{1}{n}\left[e^{2}\left\{\frac{\left(e^{-3}\right)-1}{\left(e^{-\frac{3}{n}}\right)-1}\right\}\right]\)
\(=e^{2} \cdot\left(e^{-3}-1\right) \lim _{n \rightarrow \infty} \frac{1}{n} \cdot\left(-\frac{n}{3}\right)\left[\left\{\frac{-\frac{3}{n}}{\left(e^{-\frac{3}{n}}\right)-1}\right\}\right]\)
\(=-\frac{\left(e^{2} \cdot\left(e^{-3}-1\right)\right)}{3} \lim _{n \rightarrow \infty}\left[\left\{\frac{-\frac{3}{n}}{\left(e^{-\frac{3}{n}}\right)-1}\right\}\right]\)
\(\text {as, } \lim _{n \rightarrow \infty}\left[\frac{x}{\left(e^{x}\right)-1}\right]=1\)
\(=\frac{-e^{-1}+e^{2}}{3}(1)\)
\(\Rightarrow \mathrm{I}=\frac{1}{3}\left(e^{2}-\frac{1}{e}\right)\)

Given: \( \int \frac{d x}{e^{x}+e^{-x}} \)
Let \( \mathrm{I}=\int \frac{d x}{e^{x}+e^{-x}} \)
\(=\int \frac{e^{x} d x}{\left(e^{2 x}+1\right)}\)
Put \( \mathrm{e}^{x}=\mathrm{t} \Rightarrow \mathrm{e}^{x} \mathrm{dx}=\mathrm{dt} \)
\(\Rightarrow \int \frac{e^{x} d x}{\left(e^{2 x}+1\right)}=\int \frac{d t}{\left(t^{2}+1\right)}\)
\(=\tan ^{-1} t+C\)
\(=\tan ^{-1}\left(e^{x}\right)+C\)
Hence, correct option is (A).

Given: \( \int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x \)
Let \( \mathrm{I}=\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x \)
\( =\int \frac{\cos ^{2} x-\sin ^{2} x}{(\sin x+\cos x)^{2}} d x \)
\( =\int \frac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^{2}} d x \)
\( =\int \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x \)
Put \( \sin x+\cos x =\mathrm{t} \Rightarrow \cos x -\sin x =\mathrm{dt} \)
\(\Rightarrow \int \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x=\int \frac{d t}{t}\)
\(=\log |t|+C\)
\(=\log |\sin x+\cos x|+C\)
Hence, correct option is (B).

Given: \( \int_{a}^{b}x \mathrm{f}(x) d x \)
Let \( \mathrm{I}=\int_{a}^{b} x\mathrm{f}(x) d x \)
as, \( \{\mathrm{f}(x)=\mathrm{f}(\mathrm{a}+\mathrm{b}-x)\} \)
\(\Rightarrow \mathrm{I}=\int_{a}^{b}(\mathrm{a}+\mathrm{b}-x) \mathrm{f}(\mathrm{a}+\mathrm{b}-x) \mathrm{d}x\)
\(\Rightarrow \mathrm{I}=\int_{a}^{b}(\mathrm{a}+\mathrm{b}-x) \mathrm{f}(x) \mathrm{d}x\)
\( \Rightarrow \mathrm{I}=\int_{a}^{b}(\mathrm{a}+\mathrm{b}-x) \mathrm{f}(x) \mathrm{d}x\)
\(\Rightarrow \mathrm{I}=\int_{a}^{b}(\mathrm{a}+\mathrm{b}-x) \mathrm{f}(x) \mathrm{d}x-\int_{a}^{b}(x) \mathrm{f}(x) \mathrm{d}x\)
\(\Rightarrow \mathrm{I}=\int_{a}^{b}(\mathrm{a}+\mathrm{b}-x) \mathrm{f}(x) \mathrm{d}x-\mathrm{I}\)
\(\Rightarrow 2 \mathrm{I}=\int_{a}^{b}(\mathrm{a}+\mathrm{b}) \mathrm{f}(x) \mathrm{d}x\)
\(\Rightarrow \mathrm{I}=\frac{(a+b)}{2} \int_{a}^{b} \mathrm{f}(x) d x \)
Hence, correct option is (D).

Given: \( \int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x \)
Let \( \mathrm{I}=\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x \)
\(=\int_{0}^{1} \tan ^{-1}\left(\frac{x+x-1}{1+x(1-x)}\right) d x\)
\(=\int_{0}^{1} \tan ^{-1}\left(\frac{x-(1-x)}{1+x(1-x)}\right) d x\)
\(\text {as, } \tan ^{-1}\left(\frac{A-B}{(1+A B)}\right)=\tan ^{-1}(A) \tan ^{-1}(B)\)
\(=\int_{0}^{1}\left[\tan ^{-1}(x)-\tan ^{-1}(1-x)\right] d x\quad\ldots \text{(1)}\)
\(\text {as, }\left\{\int_{0}^{a} \mathrm{f}(x) \mathrm{d}x=\int_{0}^{a} \mathrm{f}(\mathrm{a}-x) \mathrm{d}x\right\}\)
\(=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(1-(1-x))\right] d x\)
\(=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(x)\right] d x\quad\ldots \text{(2)}\)
Adding (1) and (2), we get
\(2 \mathrm{I}=\int_{0}^{1}\left[\tan ^{-1}(x)-\tan ^{-1}(1-x)\right] d x+\int_{0}^{1}\left[\tan ^{-1}(1-x)\right.\)\(\left.-\tan ^{-1}(x)\right] d x\)
\(2 \mathrm{I}=\int_{0}^{1}\left[\tan ^{-1}(x)-\tan ^{-1}(1-x)+\tan ^{-1}(1-x)-\tan ^{-1}(x)\right] d x\)
\(\Rightarrow 2 \mathrm{I}=0\)
\(\Rightarrow \mathrm{I}=0\)
Hence, correct option is (B).