Ex 7.9 Class 12 Maths Ncert Solutions

\(\text { Let } \mathrm{I}=\int_{-1}^{1}(x+1) d x\)
\(\therefore \mathrm{I}=\int_{-1}^{1}(x+1) d x\)
\(\Rightarrow \mathrm{I}=\int_{-1}^{1} x d x+\int_{-1}^{1} 1 \times d x\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]\)
\(\Rightarrow \mathrm{I}=\left[\frac{x^{2}}{2}\right]_{-1}^{1}+[x]_{-1}^{1}\)
\(\Rightarrow \mathrm{I}=\left[\frac{1^{2}}{2}-\frac{(-1)^{2}}{2}\right]+[1-(-1)]\)
\(\Rightarrow \mathrm{I}=\left[\frac{1}{2}-\frac{1}{2}\right]+[1+1]=0+2\)
\(\Rightarrow \mathrm{I}=2\)
\(\therefore \int_{-1}^{1}(x+1) d x=2\)

Let \( \mathrm{I}=\int_{2}^{3} \frac{1}{x} d x \)
\(\Rightarrow \mathrm{I}=\int_{2}^{3} \frac{1}{x} d x\left[\int {\frac{1}{x}} d x=\log x\right]\)
\(\Rightarrow \mathrm{I}=[\log |x|]_{2}^{3}\)
\(\Rightarrow \mathrm{I}=\log |3|-\log |2|\)
\(\Rightarrow \mathrm{I}=\log \frac{ 3 }{ 2 }\)
\(\therefore \int_{2}^{3} \frac{1}{x} d x=\log \frac{3}{2}\)

\(\text {Let } \mathrm{I}=\int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x\)
\(\Rightarrow \mathrm{I}=\int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x\)
\(\Rightarrow \mathrm{I}=\int_{1}^{2} 4 x^{3} d x-\int_{1}^{2} 5 x^{2} d x+\int_{1}^{2} 6 x d x+\int_{1}^{2} 9 d x\)
\(\Rightarrow \mathrm{I}=4 \int_{1}^{2} x^{3} d x-5 \int_{1}^{2} x^{2} d x+6 \int_{1}^{2} x d x+9 \int_{1}^{2} d x\)
\(\Rightarrow \mathrm{I}=4 \times\left[\frac{x^{3+1}}{3+1}\right]_{1}^{2}-5 \times\left[\frac{x^{2+1}}{2+1}\right]_{1}^{2}+6 \times\left[\frac{x^{1+1}}{1+1}\right]_{1}^{2}+9 \times\left[\frac{x^{0+1}}{0+1}\right]_{1}^{2}\)\(\left[x^{n} d x=\frac{x^{n+1}}{n+1}\right]\)
\(\Rightarrow \mathrm{I}=4 \times\left[\frac{x^{4}}{4}\right]_{1}^{2}-5 \times\left[\frac{x^{2}}{3}\right]_{1}^{2}+6 \times\left[\frac{x^{2}}{2}\right]_{1}^{2}+9 \times[x]_{1}^{2}\)
\(=2^{4}-1^{4}-5\left[\frac{2^{3}}{3}-\frac{1^{3}}{3}\right]+6\left[\frac{2^{2}}{2}-\frac{1^{2}}{2}\right]+9 \times[2-1]\)
\(=16-1-5\left[\frac{7}{3}\right]+3(3)+9\)
\(=33-\frac{35}{3}\)
\(=\frac{99-35}{3}=\frac{64}{3}\)

Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \sin 2 x d x \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \sin 2 x d x\)
\(\Rightarrow \mathrm{I}=\left[-\frac{\cos 2 x}{2}\right]_{0}^{\frac{\pi}{4}}\left[\int \sin x d x=-\cos x\right]\)
\(\Rightarrow \mathrm{I}=-\frac{(cos 2 \times \frac{ \pi }{ 4 } – cos 0)}{2}\)
\(\Rightarrow \mathrm{I}=- \frac{(\cos\frac{ \pi }{ 2 }-\cos 0)}{2}=-\frac{ (0-1) }{ 2 }\)
\(\Rightarrow \mathrm{I}=\frac{ 1 }{ 2 }\)
\(\therefore \int_{0}^{\frac{\pi}{4}} \sin 2 x d x=\frac{ 1 }{ 2 }\)

\(\text { Let } \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \cos 2 x d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \cos 2 x d x\)
\(\Rightarrow \mathrm{I}=\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}\left[\int \cos x d x=\sin x+c\right]\)
\(\Rightarrow \mathrm{I}=\frac{1}{2}\left(\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right)\)
\(\Rightarrow \mathrm{I}=\frac{1}{2}(\sin \pi-\sin \theta)\)
\(\Rightarrow \mathrm{I}=\frac{ 1 }{ 2 } \times(0-0)=0\)
\(\therefore \int_{0}^{\frac{\pi}{2}} \cos 2 x d x=0\)

\(\text {Let } \mathrm{I}=\int_{4}^{5} e^{x} d x\)
\(\Rightarrow \mathrm{I}=\int_{4}^{5} e^{x} d x\)
\(\Rightarrow \mathrm{I}=\left[e^{x}\right]_{4}^{5}=\mathrm{e}^{5}-\mathrm{e}^{4}\left[e^{x} d x=e^{x}+c\right]\)
\(\Rightarrow \mathrm{I}=\mathrm{e}^{4}(\mathrm{e}-1)\)
\(\therefore \int_{4}^{5} e^{x} d x=\mathrm{e}^{4}(\mathrm{e}-1)\)

\(\text { Let } \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \tan x d x\)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \tan x d x\left[\int \tan x d x=-\log |\cos x|+c\right]\)
\(\Rightarrow \mathrm{I}=[-\log |\cos x|]_{0}^{\frac{\pi}{4}}=-\left(\log \left|\cos \frac{\pi}{4}\right|-\log |\cos \theta|\right)\)
\(\Rightarrow \mathrm{I}=-\left(\log \left|\frac{1}{\sqrt{2}}\right|-\log |1|\right)=-\log (2)^{-\frac{1}{2}}+0\)
\(\Rightarrow \mathrm{I}=\frac{1}{2} \log 2\)
\(\therefore\int_{0}^{\frac{\pi}{4}} \tan x d x=\frac{1}{2} \log 2\)

Let \( \mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x \)
\(\Rightarrow \mathrm{I}=[\log |\operatorname{cosec} x-\cot x|]_{\frac{\pi}{6}}^{\frac{\pi}{4}}[\operatorname{cosec} x d x=\log |\operatorname{cosec} x-\cot x|+c]\)
\(\Rightarrow \mathrm{I}=\log |\operatorname{cosec} \frac{ \pi }{ 4 }-\cot \frac{ \pi }{ 4 }|-\log |\operatorname{cosec} \frac{ \pi }{ 6 }-\cot\frac{ \pi }{ 6 }|\)
\(\Rightarrow \mathrm{I}=\log |\sqrt{2}-1|-\log |2-\sqrt{3}|\)
\(\Rightarrow \mathrm{I}=\log \left|\frac{\sqrt{2}-1}{2-\sqrt{3}}\right|\)
\(\therefore \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\)

Let \( \mathrm{I}=\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}\left[\frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+c\right]\)
\(\Rightarrow \mathrm{I}=\left[\sin ^{-1} x\right]_{0}^{1}\)
\(\Rightarrow \mathrm{I}=\sin ^{-1}(1)-\sin ^{-1}(0)=\frac{ \pi }{ 2-0 }\)
\(\Rightarrow \mathrm{I}=\frac{ \pi }{ 2 }\)
\(\therefore \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}=\frac{\pi}{2}\)

Let \( \mathrm{I}=\int_{0}^{1} \frac{d x}{1+x^{2}} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{1} \frac{d x}{1+x^{2}}\)
We know that,
\(\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c\)
Therefore,
\(\Rightarrow \mathrm{I}=\left[\tan ^{-1} x\right]_{0}^{1}\)
\(\Rightarrow \mathrm{I}=\tan ^{-1}(1)-\tan ^{-1}(0)=\frac{ \pi }{ 4-0 }\)
\(\Rightarrow \mathrm{I}=\frac{ \pi }{ 4 }\)
\(\therefore \int_{0}^{1} \frac{d x}{1+x^{2}}=\frac{ \pi }{ 4 }\)

\(\text { Let } \mathrm{I}=\int_{2}^{3} \frac{d x}{x^{2}-1}\)
\(\Rightarrow \mathrm{I}=\int_{2}^{3} \frac{d x}{x^{2}-1}\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \frac{x-a}{x+a}+c\right]\)
\(\Rightarrow \mathrm{I}=\left[\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|\right]_{2}^{3}=\frac{1}{2}\left(\log \left|\frac{3-1}{3+1}\right|-\log \left|\frac{2-1}{2+1}\right|\right)\)
\(\Rightarrow \mathrm{I}=\left(\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\right)=\frac{1}{2} \log \frac{\frac{1}{2}}{\frac{1}{3}}\)
\(\Rightarrow \mathrm{I}=\frac{1}{2} \log \frac{ 3 }{ 2 }\)
\(\therefore \int_{2}^{3} \frac{d x}{x^{2}-1}= \frac{1}{2}\log \frac{ 3 }{ 2 }\)

Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x \)
\( \cos 2 x=2 \cos ^{2} x-1 \)
\( \cos ^{2} x=\frac{1+\cos 2 x}{2} \)
putting the value \( \cos ^{2} x\) in I
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} d x+\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x\)\(\left[\int \cos x d x=\sin x+c\right]\)
\(\Rightarrow \mathrm{I}=[x]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}=\frac{1}{2}\left(\frac{\pi}{2}-0\right)+\frac{1}{4}\left(\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right)\)
\(\Rightarrow \mathrm{I}=\frac{\pi}{4}+\frac{1}{4}=\frac{ \pi }{ 4 }\)
\(\therefore \int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x=\frac{ \pi }{ 4 }\)

Let \( \mathrm{I}=\int_{2}^{3} \frac{x d x}{x^{2}+1} \)
Let \( x^{2}+1=\mathrm{t} \ldots \) (i)
\(\therefore \mathrm{d}\left(x^{2}+1\right)=\mathrm{dt}\)
\(\Rightarrow 2 x \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow x\mathrm{d}x=\frac{ \mathrm{dt} }{ 2 }\)
When \( x=2 ; \mathrm{t}=2^{2}+1=5 \)
When \( x=3 ; \mathrm{t}=3^{2}+1=10 \)
Substituting \( x^{2}+1 \) and \(xdx\) in I
\(\Rightarrow \mathrm{I}=\int_{5}^{10} \frac{d t}{2 t}=\frac{1}{2} \int_{5}^{10} \frac{d t}{t}\left[\int \frac{1}{x} d x=\log x\right]\)
\(\Rightarrow \mathrm{I}=\frac{1}{2}[\log t]_{5}^{10}=\frac{1}{2}(\log 10-\log 5)=\frac{1}{2} \log \frac{10}{5}\)
\(\Rightarrow \mathrm{I}=\frac{ 1 }{ 2 }\log 2\)
\(\int_{2}^{3} \frac{x d x}{x^{2}+1}=\frac{ 1 }{ 2 }\log 2\)

Let \( \mathrm{I}=\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} \)
Multiplying by 5 in numerator and denominator
\(\Rightarrow \mathrm{I}=\frac{1}{5} \int_{0}^{1} \frac{5(2 x+3)}{5 x^{2}+1} d x=\frac{1}{5} \int_{0}^{1} \frac{10 x+15}{5 x^{2}+1} d x\)
\(\Rightarrow \mathrm{I}=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x+3 \int_{0}^{1} \frac{1}{5 x^{2}+1}\)
\(\Rightarrow \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}\)
\(\mathrm{I}_{1}=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x\)
Let \( 5 x^{2}+1=\mathrm{t} \ldots \) (i)
\( \mathrm{d}\left(5 x^{2}+1\right)=\mathrm{dt} \)
\( 10 x \mathrm{dx}=\mathrm{dt} \quad\ldots \text{(ii)}\)
When \( x=0 ; \mathrm{t}=5 \times 02+1=1 \)
When \( x=1 ; \mathrm{t}=5 \times 12+1=6 \)
Substituting (i) and (ii) in I1
\(\mathrm{I}_{1}=\frac{1}{5} \int_{1}^{6} \frac{d t}{t}=\frac{1}{5}[\log |t|]_{1}^{6}\left[\frac{1}{x} d x=\log x\right]\)
\(\mathrm{I}_{1}=\frac{1}{5}(\log |6|-\log |1|)=\frac{1}{5}(\log 6-0)\)
\(\mathrm{I}_{1}=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x=\frac{\log 6}{5}\)
\(\mathrm{I}_{2}=\frac{3}{5} \times \frac{1}{\frac{1}{\sqrt{5}}}\left[\tan ^{-1} \sqrt{5} x\right]_{0}^{1}=\frac{3}{5} \times \sqrt{5}\left(\tan ^{-1} \sqrt{5} \tan ^{-1} 0\right)\)
\(\mathrm{I}_{2}=\frac{ 3 }{ \sqrt{ 5} } \tan ^{-1} 5\)
\(\because \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}\)
\(\therefore \mathrm{I}=\frac{ 1 }{ 5 }\log 6+\frac{ 3 }{ \sqrt{ } 5 } \tan ^{-1} 5\)
\(\therefore \int_{0}^{1} \frac{2 x+3}{5 x^{2}+1}=\frac{ 1 }{ 5 } \log 6+\frac{ 3 }{ \sqrt{ 5 }} \tan ^{-1} 5\)

Let \( \mathrm{I}=\int_{0}^{1} x e^{x^{2}} d x \)
Put \( x^{2}=\mathrm{t} \Rightarrow 2 x\mathrm{d}x=\mathrm{dt} \)
When \( x=0 ; \mathrm{t}=0 \)
When \( x=1 ; \mathrm{t}=1 \)
Substituting \( t \) and dt in I
\(\Rightarrow \mathrm{I}=\int_{0}^{1} \frac{e^{t} d t}{2}=\frac{ 1 }{ 2 }\int_{0}^{1} e^{t} d t\left[\int e^{x} d x=e^{x}+c\right]\)
\(\Rightarrow \mathrm{I}=\frac{1}{2}\left[e^{t}\right]_{0}^{1}=\frac{1}{2}\left(e-e^{0}\right)=\frac{1}{2}(e-1)\)
\(\therefore \int_{0}^{1} x e^{x^{2}} d x=\frac{ 1 }{ 2 }(\mathrm{e}-1)\)

Let \( \mathrm{I}=\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} \)
Dividing \( 5 x^{2} \) by \( x^{2}+4 x+3 \) we get 5 as quotient and \( -(20 x+15) \) as remainder
\(\text { So, } \mathrm{I}=\int_{1}^{2}\left(5-\frac{20 x+15}{x^{2}+4 x+3}\right) d x\)
\(\Rightarrow \mathrm{I}=\int_{1}^{2} 5 d x-\int_{1}^{2} \frac{20 x+15}{x^{2}+4 x+3}=5[x]_{1}^{2}-\int_{1}^{2} \frac{20 x+15}{x^{2}+4 x+3}\)
\(\Rightarrow \mathrm{I}=5(2-1)-\int_{1}^{2} \frac{20 x+15}{x^{2}+4 x+3}\)
\(\Rightarrow \mathrm{I}=5-\mathrm{I}_{1}\)
\(\mathrm{I}_{1}=\int_{1}^{2} \frac{20 x+15}{x^{2}+4 x+3}\)
Adding and subtracting 25 in the numerator
\(\mathrm{I}_{1}=\int_{1}^{2} \frac{20 x+15+25-25}{x^{2}+4 x+3} d x=\int_{1}^{2} \frac{20 x+40}{x^{2}+4 x+3} d x-\int_{1}^{2} \frac{25}{x^{2}+4 x+3} d x\)
\(\mathrm{I}_{1}=10 \int_{1}^{2} \frac{20 x+4}{x^{2}+4 x+3} d x-25 \int_{1}^{2} \frac{1}{x^{2}+4 x+3} d x\)
Let \( x^{2}+4 x+3=\mathrm{t} \)
\((2 x+4) \mathrm{d}x=\mathrm{dt}\)
\(\therefore \mathrm{I}_{1}=10 \int \frac{d t}{t}-25 \int_{1}^{2} \frac{1}{x^{2}+4 x+3+1-1} d x=10 \log t+25 \int_{1}^{2} \)\(\frac{1}{x^{2}+4 x+4-1} d x\)
\(\mathrm{I}_{1}=10 \log \mathrm{t}-25 \int_{1}^{2} \frac{1}{(x+2)^{2}-1^{2}} d x\left[\int \frac{1}{x} d x=\log x\right]\)
\(\mathrm{I}_{1}=10 \log \mathrm{t}-25\left[\frac{1}{2} \log \left(\frac{x+2-1}{x+2+1}\right)\right]\left[\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \frac{x-a}{x+a}+c\right]\)
\(\mathrm{I}_{1}=10\left[\log \left(x^{2}+4 x+3\right)\right]_{1}^{2}-\frac{25}{2}\left[\log \left(\frac{x+1}{x+3}\right)\right]_{1}^{2}\)
\(\mathrm{I}_{1}=10\left[\log \left(2^{2}+4 \times 2+3\right)-\log \left(1^{2}+4 \times 1+3\right)\right]-\frac{25}{2}\)\(\left[\log \left(\frac{2+1}{2+3}\right)-\log \left(\frac{1+1}{1+3}\right)\right]\)
\(\mathrm{I}_{1}=10[\log 15-\log 8]-\frac{25}{2}\left[\log \frac{3}{5}-\log \frac{2}{4}\right]\)
\(\mathrm{I}_{1}=10[\log (5 \times 3)-\log (4 \times 2)]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]\)
\(\mathrm{I}_{1}=10 \log 5+10 \log 3-10 \log 4-10 \log 2 -\frac{25}{2} \log 3+\frac{25}{2} \log 5+\)\(\frac{25}{2} \log 2-\frac{25}{2} \log 4 \)
\(\mathrm{I}_{1}=\left(10+\frac{25}{2}\right) \log 5-\left(10+\frac{25}{2}\right) \log 4+\left(10-\frac{25}{2}\right) \log 3+\)\(\left(-10+\frac{25}{2}\right) \log 2\)
\(\mathrm{I}_{1}=\frac{45}{2} \log 5-\frac{45}{2} \log 4-\frac{5}{2} \log 3+\frac{5}{2} \log 2=\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\)
\(\because \mathrm{I}=5-\mathrm{I}_{1}\)
Substituting \( I_{1} \) in I we get
\(\mathrm{I}=5-\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\)
\(\therefore \int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3}=5-\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\)

Let \( \mathrm{I}=\int_{0}^{\frac{\pi}{4}}\left(2 \sec ^{2} x+x^{3}+2\right) d x \)
\(\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}}\left(2 \sec ^{2} x+x^{3}+2\right) d x=2 \int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x+\int_{0}^{\frac{\pi}{4}} x^{3} \)\(d x+2 \int_{0}^{\frac{\pi}{4}} d x\)
\(\Rightarrow \mathrm{I}=2[\tan x]_{0}^{\frac{\pi}{4}}+\left[\frac{x^{4}}{4}\right]_{0}^{\frac{\pi}{4}}+2[x]_{0}^{\frac{\pi}{4}}\left[\int \sec ^{2} x d x=\tan x+c\right]\)
\(\Rightarrow \mathrm{I}=2(\tan \frac{ \pi }{ 4 }-\tan 0)+\left((\frac{ \pi }{ 4 })^{4}-0\right)+2(\frac{ \pi }{ 4-0 })\)
\(\Rightarrow \mathrm{I}=2 \times 1+\frac{1}{4} \times\left(\frac{\pi}{4}\right)^{4}+2 \times \frac{\pi}{4}\)
\(\Rightarrow \mathrm{I}=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}\)
\(\therefore \int_{0}^{\frac{\pi}{4}}\left(2 \sec ^{2} x+x^{3}+2\right) d x=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}\)

Let \( \mathrm{I}=\int_{0}^{\pi}\left(\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}\right) d x \)
We know, \( \cos x=\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2} \)
Substituting \( \cos x=\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2} \) in I, we get
\(\Rightarrow \mathrm{I}=\int_{0}^{\pi} \cos x d x=[\sin x]_{0}^{\pi} \quad\left[\int \cos x d x=\sin x+c\right]\)
\(\Rightarrow \mathrm{I}=\sin \pi-\sin 0=0-0=0\)
\(\therefore \int_{0}^{\pi}\left(\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}\right) d x=0\)

Let \( \mathrm{I}=\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x \)
\(\mathrm{I}=3 \int_{0}^{2} \frac{2 x+1}{x^{2}+4}=3 \int_{0}^{2} \frac{2 x}{x^{2}+4} d x+3 \int_{0}^{2} \frac{1}{x^{2}+4} d x\)
\(\therefore \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}\)
\(\mathrm{I}_{1}=3 \int_{0}^{2} \frac{1}{x^{2}+4} d x\)
Let \( x^{2}+4=\mathrm{t} \)
\(2 x\mathrm{d}x=\mathrm{dt}\)
When \( x=0 ; \mathrm{t}=4 \)
When \( x=2 ; \mathrm{t}=2^{2}+4=8 \)
Substituting t and dt in \( \mathrm{I}_{1} \)
\(\Rightarrow \mathrm{I}_{1}=3 \int_{4}^{8} \frac{d t}{t}=3[\log |t|]_{4}^{8} \quad\left[\int \frac{1}{x} d x=\log x\right]\)
\(\Rightarrow \mathrm{I}_{1}=3[\log |8|-\log |4|]=3 \log \frac{ 8 }{ 4 }\)
\(\Rightarrow \mathrm{I}_{1}=3 \log 2=-3 \log 2\)
\(\mathrm{I}_{2}=3 \int_{0}^{2} \frac{1}{x^{2}+4} d x=3 \int_{0}^{2} \frac{1}{x^{2}+2^{2}} d x\left[\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c\right]\)
\(\Rightarrow \mathrm{I}_{2}=3 \times \frac{1}{2}\left[\tan ^{-1} \frac{x}{2}\right]_{0}^{2}=\frac{3}{2}\left[\tan ^{-1} \frac{2}{2}-\tan ^{-1} \frac{0}{2}\right]\)\(=\frac{3}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]\)
\(\Rightarrow \mathrm{I}_{2}=\frac{3}{2} \times \frac{\pi}{4}= \frac{ 3\pi }{ 8 }\)
Now I \( =\mathrm{I}_{1}+\mathrm{I}_{2} \)
\( \mathrm{I}=3 \log 2+ \frac{3 \pi }{ 8 } \)
\(\therefore \int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x=3 \log 2 + \frac{ 3\pi }{ 8 }\)

Let \( \mathrm{I}=\int_{0}^{1}\left(x e^{x}+\sin \frac{\pi x}{4}\right) d x \)
\(\Rightarrow \mathrm{I}=\int_{0}^{1} x e^{x} d x+\int_{0}^{1} \sin \frac{\pi x}{4} d x\)
\(\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}\)
\(\mathrm{I}_{1}=x \int e^{x} d x-\int\left\{\left(\frac{d}{d x} x\right) \int e^{x} d x\right\} d x\)
\(\Rightarrow \mathrm{I}_{1}=x e^{x}-\int e^{x} d x \quad\left[\int e^{x} d x=e^{x}+c\right]\)
\(\Rightarrow \mathrm{I}_{1}=\left[x e^{x}-e^{x}\right]_{0}^{1}=\left[\left(1 \times e^{1}-e^{1}\right)-\left(0 \times e^{0}-e^{0}\right)\right]\)
\(\Rightarrow \mathrm{I}_{1}=\mathrm{e}-\mathrm{e}-0+1\)
\(\Rightarrow \mathrm{I}_{1}=1\)
\(\mathrm{I}_{2}=\int_{0}^{1} \sin \frac{\pi x}{4} d x\)
\(\Rightarrow \mathrm{I}_{2}=\left[-\frac{\cos \frac{\pi x}{4}}{\frac{\pi}{4}}\right]_{0}^{1}=\frac{4}{\pi}\left[\cos \frac{\pi}{4} \times 1-\cos \frac{\pi}{4} \times 0\right]=\frac{4}{\pi}\)\(\left[\cos \frac{\pi}{4} \times \cos 0\right]\)
\(\Rightarrow \mathrm{I}_{2}=\frac{4}{\pi}\left(1-\frac{1}{\sqrt{2}}\right)=\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\)
Since, \( I=I_{1}+I_{2} \)
\(\therefore \mathrm{I}=1+\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\)
\(\therefore \int_{0}^{1}\left(x e^{x}+\sin \frac{\pi x}{4}\right) d x=1+\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\)

Let \( \mathrm{I}=\int_{1}^{\sqrt{3}} \frac{d x}{x^{2}+1} \)
\(\Rightarrow \mathrm{I}=\int_{1}^{\sqrt{3}} \frac{d x}{x^{2}+1}\)
\(\Rightarrow \mathrm{I}=\left[\tan ^{-1} x\right]_{1}^{\sqrt{3}}=\left[\tan ^{-1} \sqrt{3} \tan ^{-1} 1\right]=\frac{\pi}{3}-\frac{\pi}{4}\)\(\left[\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c\right]\)
\(\Rightarrow \mathrm{I}=\frac{4 \pi-3 \pi}{12} \frac{ \pi }{ 12 }\)
\(\therefore \int_{1}^{\sqrt{3}} \frac{d x}{x^{2}+1}=\frac{ \pi }{ 12 }\)

Let \( \mathrm{I}=\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}} \)
\(\Rightarrow \mathrm{I}=\int_{0}^{3} \frac{d x}{4+9 x^{2}}\)
Taking 9 common from Denominator in I
\(\Rightarrow \mathrm{I}=\frac{1}{9} \int_{0}^{\frac{2}{3}} \frac{d x}{\frac{4}{9}+x^{2}}=\frac{1}{9} \int_{0}^{\frac{2}{3}} \frac{d x}{\left(\frac{2}{3}\right)^{2}+x^{2}}\left[\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c\right]\)
\(\Rightarrow \mathrm{I}=\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{x}{\frac{2}{3}}\right]_{0}^{\frac{2}{3}}=\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{3 x}{2}\right]_{0}^{\frac{2}{3}}\)
\(\Rightarrow \mathrm{I}=\frac{1}{6}\left[\tan ^{-1} \frac{3}{2} \times \frac{2}{3}-\tan ^{-1} 0\right]=\frac{1}{6}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]\)
\(\Rightarrow \mathrm{I}=\frac{1}{6} \times\left(\frac{\pi}{4}-0\right)=\frac{ \pi }{ 24 }\)
\(\therefore \int_{0}^{3} \frac{d x}{4+9 x^{2}}=\frac{ \pi }{ 24 }\)