Ex 7.4 Class 12 Maths Ncert Solutions

Let \( x^{3}=\mathrm{t} \)
\(\Rightarrow 3 x^{2} \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{3 x^{2}}{x^{6}+1} d x=\int \frac{d t}{\left(t^{2}+1\right)}\)
\(=\tan ^{-1} \mathrm{t}+\mathrm{C}\)
\(=\tan ^{-1}\left(x^{3}\right)+\mathrm{C}\)

Let \( 2 x=\mathrm{t} \)
\(\Rightarrow 2 \mathrm{d}x=\mathrm{dt}\)
\(\frac{1}{\sqrt{1+4 x^{2}}}\)
\(=\int \frac{d t}{2 \sqrt{1+t^{2}}}\)
\(\Rightarrow\frac{1}{2}\left[\log \left|t+\sqrt{t^{2}+1}\right|\right]+C\)
\(\Rightarrow\frac{1}{2}\left[\log \left|2 x+\sqrt{4 x^{2}+1}\right|\right]+C\)

Let \( 2-x=\mathrm{t} \)
\(\Rightarrow-\mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{\sqrt{(2-x)^{2}+1}} d x=-\int \frac{d t}{\sqrt{t^{2}+1}}\)
\(\Rightarrow-\left[\log \left|t+\sqrt{t^{2}+1}\right|\right]+C\)
\(\Rightarrow\left[\log \left|(2-x)+\sqrt{(2-x)^{2}+1}\right|\right]+C\)
\(=\log \left|\frac{1}{(2-x)+\sqrt{x^{2}-4 x+5}}\right|+C\)

Let \( 5 x=\mathrm{t} \)
\(\Rightarrow 5 \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{\sqrt{1-25 x^{2}}} d x=-\int \frac{d t}{\sqrt{3^{2}-t^{2}}}\)
\(\Rightarrow\frac{1}{5} \sin ^{-1}\left(\frac{t}{3}\right)+C\)
\(\Rightarrow\frac{1}{5} \sin ^{-1}\left(\frac{5 x}{3}\right)+C\)

Let \( \sqrt{2} x^{2}=t \)
\(\Rightarrow 2 \sqrt{2 x} d x=\mathrm{dt}\)
\(\Rightarrow \int \frac{3 x}{1+2 x^{2}} d x=\frac{3}{2 \sqrt{2}} \int \frac{d t}{1+t^{2}}\)
\(\Rightarrow\frac{3}{2 \sqrt{2}}\left[\tan ^{-1} t\right]+C\)
\(\Rightarrow\frac{3}{2 \sqrt{2}} \tan ^{-1} \sqrt{2} x^{2}+C\)

Let \( x^{3}=\mathrm{t} \)
\(\Rightarrow 3 x^{2} \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{x^{2}}{1-x^{6}} d x=\frac{1}{3} \int \frac{d t}{1-t^{2}}\)
\(\Rightarrow\frac{1}{3}\left[\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\right]+C\)
\(\Rightarrow\frac{1}{6}\left[\log \left|\frac{1+x^{3}}{1-x^{3}}\right|\right]+C\)

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

Let \( x^{3}=\mathrm{t} \)
\(\Rightarrow 3 x^{2} \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \frac{x^{2}}{\sqrt{x^{6}+a^{6}}} d x=\frac{1}{3} \int \frac{d t}{\sqrt{t^{2}+\left(a^{3}\right)^{2}}}\)
\(\Rightarrow\frac{1}{3} \log \left|t+\sqrt{t^{2}+a^{6}}\right|+C\)
\(\Rightarrow\frac{1}{3} \log \left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+C\)

Let \( \tan x=\mathrm{t} \)
\(\Rightarrow \sec ^{2} x\mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}} d x=\int \frac{d t}{\sqrt{t^{2}+2^{2}}}\)
\(\Rightarrow\log \left|t+\sqrt{t^{2}+4}\right|+C\)
\(\Rightarrow\log \left|\tan x+\sqrt{\tan ^{2} x+4}\right|+C\)

\(\int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x\)
Let \( x+1=\mathrm{t} \)
\(\Rightarrow \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x=\int \frac{1}{\sqrt{t^{2}+1}} d t\)
\(\Rightarrow\log \left|t+\sqrt{t^{2}+1}\right|+C\)
\(\Rightarrow\log \left|(x+1)+\sqrt{(x+1)^{2}+1}\right|+C\)
\(\Rightarrow\log \left|(x+1)+\sqrt{x^{2}+2 x+2}\right|+C\)

\(\Rightarrow \int \frac{1}{9 x^{2}+6 x+5} d x=\int \frac{1}{(3 x+1)^{2}+\left(2^{2}\right)} d x\)
Let \( 3 x+1=\mathrm{t} \)
\(\Rightarrow 3 \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{(3 x+1)^{2}+(2)^{2}} d x=\frac{1}{3} \int \frac{1}{t^{2}+2^{2}} d t\)
\(\Rightarrow\frac{1}{3}\left[\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)\right]+C\)
\(\Rightarrow\frac{1}{6} \tan ^{-1}\left(\frac{3 x+1}{2}\right)+C\)

\(\int \frac{1}{\sqrt{7-6 x-x^{2}}} d x=\int \frac{1}{\sqrt{(4)^{2}-(x+3)^{2}}} d x\)
Let \(x+3=\mathrm{t} \)
\(\Rightarrow \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x=\int \frac{1}{\sqrt{(4)^{2-} t^{2}}}\)
\(\Rightarrow\sin ^{-1}\left(\frac{t}{4}\right)+C\)
\(\Rightarrow\sin ^{-1}\left(\frac{x+3}{4}\right)+C\)

\(\int \frac{1}{\sqrt{(x-1)(x-2)}} d x=\int \frac{1}{\sqrt{x^{2}-3 x+2}} d x=\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x\)
Let \( x-\frac{3}{2}=\mathrm{t} \)
\(\Rightarrow \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}} d x=\int \frac{1}{\sqrt{(t)^{2}-\left(\frac{1}{2}\right)^{2}}} d t\)
\(\Rightarrow\log \left|t+\sqrt{(t)^{2}-\left(\frac{1}{2}\right)^{2}}\right|+C\)
\(\Rightarrow\log \left|\left(x-\frac{3}{2}\right)+\sqrt{x^{2}-3 x+2}\right|+C\)

\(\int \frac{1}{\sqrt{8+3 x-x^{2}}}=\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x\)
Let \( x-\frac{3}{2}=\mathrm{t} \)
\(\Rightarrow \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x=\int \frac{1}{\sqrt{\left(\frac{41}{2}\right)^{2}-(t)^{2}}} d t\)
\(\Rightarrow\sin ^{-1}\left(\frac{t}{\frac{\sqrt{41}}{2}}\right)+C\)
\(\Rightarrow\sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+C\)

\((x-1)(x-b)\) can be written as \( x^{2}-(a+b) x+a b \).
Then, \( x^{2}-(\mathrm{a}+\mathrm{b}) x+\mathrm{ab}=x^{2}-(\mathrm{a}+\mathrm{b}) x+\frac{(a+b)^{2}}{4}-\frac{(a+b)^{2}}{4}+a b \)
\(\Rightarrow\left[x-\left(\frac{a+b}{2}\right)\right]^{2}-\frac{(a-b)^{2}}{4}\)
\(\Rightarrow \therefore \int \frac{1}{\sqrt{(x-a)(x-b)}} d x=\int \frac{1}{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\frac{(a-b)^{2}}{4}} d x\)
Let \( x=\left(\frac{a+b}{2}\right)=t \)
\(\Rightarrow \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\frac{(a-b)^{2}}{4}} d x=\int \frac{1}{\sqrt{t^{2}-\frac{(a-b)^{2}}{4}}}\)
\(\Rightarrow\log \left|t+\sqrt{t^{2}-\frac{(a-b)^{2}}{4}}\right|+C\)
\(\Rightarrow\log \left|\left\{x-\left(\frac{a+b}{2}\right)\right\}+\sqrt{(x-a)(x-b)}\right|+C\)

Let \( 4 x+1=\mathrm{A} \frac{\mathrm{d}}{\mathrm{d}x}\left(2 x^{2}+x-3\right)+\mathrm{B} \)
\(\Rightarrow 4 x+1=A(4 x+1)+B\)
\(\Rightarrow 4 x+1=4 A x+A+B\)
Now, equating the coefficients of \( x \) and constant term on both sides, we get,
\(4 \mathrm{A}=4\)
\(\Rightarrow \mathrm{A}=1\)
\(\mathrm{A}+\mathrm{B}=1\)
\(\Rightarrow \mathrm{B}=0\)
Let \( 2 x^{2}+x-3=\mathrm{t} \)
\(\Rightarrow(4 x+1) \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{4 x+1}{\sqrt{2 x^{2}+x-3}} d x=\int \frac{1}{\sqrt{t}} d t\)
\(\Rightarrow2 \sqrt{t}+\mathrm{C}\)
\(\Rightarrow \sqrt[2]{2 x^{2}+x-3}+\mathrm{C}\)

Let \( x+2=A \frac{d}{d x}(-1)+B \)
\(\Rightarrow x+2=\mathrm{A}(2 x)+\mathrm{B}\)
Now, equating the coefficients of \(x\) and constant term on both sides, we get,
\( 2 \mathrm{A}=1 \)
\( \Rightarrow \mathrm{A}=\frac{1}{2} \)
\( B=2 \)
\( \Rightarrow x+2=\frac{1}{2}(2 x)+2 \)
\( \Rightarrow \int \frac{x+2}{\sqrt{x^{2}-1}} d x=\int \frac{\frac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} \)
\( \Rightarrow\frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x+2 \int \frac{1}{\sqrt{x^{2}-1}} d x \)
Now, \( \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x \)
Let \( x^{2}-1=\mathrm{t} \)
\(\Rightarrow(2 x) \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2}-1}} d x=\frac{1}{2} \int \frac{d t}{\sqrt{t}}=\frac{1}{2}[2 \sqrt{t}]\)
\(\Rightarrow\sqrt{t}=\sqrt{x^{2}-1}\)
And \( 2 \int \frac{1}{\sqrt{x^{2}-1}} d x=2 \log \left|x+\sqrt{x^{2}-1}\right| \)
\(\Rightarrow \int \frac{x+2}{\sqrt{x^{2}-1}} d x=\sqrt{x^{2}-1}+2 \log \left|x+\sqrt{x^{2}-1}\right|+C\)

Let \( 5 x-2= \)
\(\Rightarrow 5 x-2=\mathrm{A}(2+6 x)+\mathrm{B}\)
Now, equating the coefficients of \(x\) and constant term on both sides, we get,
\(6 \mathrm{A}=5\)
\(\Rightarrow \mathrm{A}=\frac{5}{6}\)
\(2 \mathrm{A}+\mathrm{B}=-2\)
\(\Rightarrow \mathrm{B}=-\frac{11}{3}\)
\(\Rightarrow 5 x-2=\frac{5}{6}(2+6 x)+\left(-\frac{11}{3}\right)\)
\(\Rightarrow \int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^{2}} d x\)
\(\Rightarrow\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^{2}} d x-\frac{11}{3} \int \frac{1}{1+2 x+3 x^{2}} d x\)
Now, \( \int \frac{2+6 x}{1+2 x+3 x^{2}} d x \)
Let \( 1+2 x+3 x^{2}=\mathrm{t} \)
\(\Rightarrow(2+6 x) \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{2+6 x}{1+2 x+3 x^{2}} d x=\int \frac{d t}{t}=\log |t|\)
\(=\log \left|1+2 x+3 x^{2}\right| \ldots \text { (1) }\)
And \( \int \frac{1}{1+2 x+3 x^{2}} d x \)
\(1+2 x+3 x^{2}=1+3\left(x^{2}+\frac{2}{3} x\right)\)
\(\Rightarrow3\left[\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right]\)
\(\Rightarrow \int \frac{1}{1+2 x+3 x^{2}} d x=\frac{1}{3} \int \frac{1}{\left[\left(x+\frac{1}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right]} d x\)
\(\Rightarrow\frac{1}{3}\left[\frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)\right]\)
\(\Rightarrow\frac{1}{3}\left[\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)\right]\)
\(=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right) \ldots(2)\)
Thus, from (1) and (2), we get,
\(\Rightarrow \int \frac{5 x-2}{1+2 x+3 x^{2}} d x=\frac{5}{6}\left[\log \left|1+2 x+3 x^{2}\right|\right]-\frac{11}{3}\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)\right]\)\(+C\)
\(\Rightarrow\frac{5}{6}\left[\log \left|1+2 x+3 x^{2}\right|\right]-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+C\)

Let \( 6 x+7=A \frac{d}{d x}\left(x^{2}-9 x+20\right)+B \)
\(\Rightarrow 6 x+7=A(2 x-9)+B\)
Now, equating the coefficients of \( x \) and constant term on both sides, we get,
\(2 A=6\)
\(\Rightarrow A=3\)
\(-9 A+B=7\)
\(\Rightarrow B=34\)
\(\Rightarrow 6 x+7=3(2 x-9)+34\)
\(\Rightarrow \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x=\int \frac{3(2 x-9)+34}{\sqrt{x^{2}-9 x+20}} d x\)
\(\Rightarrow3 \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x=\int \frac{1}{\sqrt{x^{2}-9 x+20}} d x\)
Now, \( \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x \)
Let \( x^{2}-9 x+20=\mathrm{t} \)
\(\Rightarrow(2 x-9) \mathrm{d}x=\mathrm{dt}\)
\(\therefore \int \frac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}=2 \sqrt{x^{2}-9 x+20}\quad\ldots \text{(1)}\)
And \( \int \frac{1}{\sqrt{x^{2}-9 x+20}} d x \)
\(x^{2}-9 x+20=x^{2}-9 x+20+\frac{81}{4}-\frac{81}{4}\)
\(\Rightarrow\left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\)
\(\Rightarrow \int \frac{1}{\sqrt{x^{2}-9 x+20}} d x=\int \frac{1}{\left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d x\)
\(=\log \left|\left(x-\frac{9}{2}\right)^{2}+\sqrt{x^{2}-9 x+20}\right|\quad \ldots(2)\)
Thus, from (1) and (2), we get,
\(\Rightarrow \int \frac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x=3\left[2 \sqrt{x^{2}-9 x+20}\right]+34 \log \left[\left(x-\frac{9}{2}\right)^{2}\right.\)\(\left.+\sqrt{x^{2}-9 x+20}\right]+C\)
\(\Rightarrow6 \sqrt{x^{2}-9 x+20}+34 \log \left[\left(x-\frac{9}{2}\right)^{2}+\sqrt{x^{2}-9 x+20}\right]+C\)

Let \( x+2=A \frac{d}{d x}\left(4 x-x^{2}\right)+B \)
\(\Rightarrow x+2=\mathrm{A}(4-2 x)+\mathrm{B}\)
Now, equating the coefficients of \( x \) and constant term on both sides, we get,
\(-2 \mathrm{A}=1\)
\(\Rightarrow \mathrm{A}=-\frac{1}{2}\)
\(4 \mathrm{A}+\mathrm{B}=2\)
\(\Rightarrow \mathrm{B}=4\)
\(\Rightarrow x+2=-\frac{1}{2}(4-2 x)+4\)
Now, \( \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} \)
\(\Rightarrow\frac{1}{2} \int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x+4 \int \frac{1}{\sqrt{4 x-x^{2}}} d x\)
Now, let us consider, \( \int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x \)
Let \( 4 x-x^{2}=\mathrm{t} \)
\(\Rightarrow(4-2 x) \mathrm{d}x=\mathrm{dt}\)
\(\therefore \int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}=2 \sqrt{4 x-x^{2}} \quad\ldots\)(1)
And. Now let us consider,
Then, \( 4 x-x^{2}=-\left(-4 x+x^{2}\right) \)
\(=\left(-4 x+x^{2}+4-4\right)\)
\(=4-(x-2)^{2}\)
\(=(2)^{2}-(x-2)^{2}\)
\(\therefore \int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x=\sin ^{-1}\left(\frac{x-2}{2}\right) \quad\ldots \)(2)
using eq. (1) and (2), we get,
\(\Rightarrow \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=\sqrt{4 x-x^{2}}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+C\)

\(\int \frac{x+2}{\sqrt{x^{2}+2 x+3}} d x=\frac{1}{2} \int \frac{2(x+2)}{\sqrt{x^{2}+2 x+3}} d x\)
\(\Rightarrow \frac{1}{2} \int \frac{2 x+4}{\sqrt{x^{2}+2 x+3}} d x\)
\(\Rightarrow \frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\frac{1}{2} \int \frac{2}{\sqrt{x^{2}+2 x+3}} d x\)
\(\Rightarrow \frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\int \frac{1}{\sqrt{x^{2}+2 x+3}} d x\)
Now. Let us consider \( \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x \)
Let \( x^{2}+2 x+3=\mathrm{t} \)
\(\Rightarrow(2 x+2) \mathrm{d}x=\mathrm{dt}\)
\(\therefore \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x=2 \sqrt{t}=2 \sqrt{x^{2}+2 x+3} \quad\ldots\)(1)
And, now let us consider \( \int \frac{1}{\sqrt{x^{2}+2 x+3}} d x \)
\(\Rightarrow x^{2}+2 x+3=x^{2}+2 x+1+2=(x+1)^{2}+(\sqrt{2})^{2}\)
\(\Rightarrow \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x=\log \left|(x+1)+\sqrt{x^{2}+2 x+3}\right|\)
Using eq. (1) and (2), we get,
\(\Rightarrow \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x=\frac{1}{2}\left[2 \sqrt{x^{2}+2 x+3}\right]+\log \left|(x+1)+\sqrt{x^{2}+2 x+3}\right|\)\(+C\)
\(\Rightarrow\sqrt{x^{2}+2 x+3}+\log \left|(x+1)+\sqrt{x^{2}+2 x+3}\right|+C\)

Let \( x+3=\mathrm{A} \frac{d}{d x}\left(x^{2}-2 x-5\right)+\mathrm{B} \)
\(\Rightarrow x+3=\mathrm{A}(2 x-2)+\mathrm{B}\)
Now, equating the coefficients of \( x \) and constant term on both sides, we get,
\(2 \mathrm{A}=1\)
\(\Rightarrow \mathrm{A}=\frac{1}{2}\)
\(-2 \mathrm{A}+\mathrm{B}=3\)
\(\Rightarrow \mathrm{B}=4\)
\(\Rightarrow x+3=\frac{1}{2}(2 x-2)+4\)
Now, \( \int \frac{x+3}{x^{2}-2 x-5} d x=\int \frac{\frac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x \)
\(\Rightarrow\frac{1}{2} \int \frac{2 x-2}{x^{2}-2 x-5} d x+4 \int \frac{1}{x^{2}-2 x-5} d x\)
Now, let us consider \( \int \frac{2 x-2}{x^{2}-2 x-5} d x \)
Let \( x^{2}-2 x-5=\mathrm{t} \)
\(\Rightarrow(2 x-2) \mathrm{d}x=\mathrm{dt}\)
\(\therefore \int \frac{2 x-2}{x^{2}-2 x-5} d x=\int \frac{d t}{t}=\log |t|=\log \left|x^{2}-2 x-5\right|\quad\ldots\)(1)
And, now let us consider, \( \int \frac{1}{x^{2}-2 x-5} d x \)
\(\Rightarrow\int \frac{1}{\left(x^{2}-2 x+1\right)-6} d x\)
\(\Rightarrow\int \frac{1}{(x-1)^{2}+(\sqrt{6})^{2}} d x\)
\(=\frac{1}{2 \sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right) \quad\ldots\) (2)
Using eq. (1) and (2), we get,
\(\Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{4}{2 \sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)+C\)
\(\Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)+C\)

Let \( 5 x+3=\mathrm{A} \frac{d}{d x}\left(x^{2}+4 x+10\right)+\mathrm{B} \)
\(\Rightarrow 5 x+3=A(2 x+4)+B\)
Now, equating the coefficients of \( x \) and constant term on both sides, we get,
\(2 \mathrm{A}=5\)
\(\Rightarrow \mathrm{A}=\frac{5}{2}\)
\(4 \mathrm{A}+\mathrm{B}=3\)
\(\Rightarrow \mathrm{B}=-7\)
\(\Rightarrow 5 x+3=\frac{5}{2}(2 x+4)-7\)
Now, \( \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} \)
\(\Rightarrow\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x\)
Now, let us consider, \( \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x \)
Let \( x^{2}+4 x+10=\mathrm{t} \)
\(\Rightarrow(2 x+4) \mathrm{d}x=\mathrm{dt}\)
\(\therefore \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{d t}{t}=2 \sqrt{t}=2 \sqrt{x^{2}+4 x+10} \quad\ldots\)(1)
And, now let us consider, \( \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x \)
\(\Rightarrow \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{1}{\left(\sqrt{x^{2}+4 x+4}\right)+6} d x\)
\(\Rightarrow\int \frac{1}{(x+2)^{2}+(\sqrt{6})^{2}} d x\)
\(=\log \left|(x+2) \sqrt{x^{2}+4 x+10}\right|\quad \ldots(2)\)
using eq. (1) and (2), we get,
\(\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\frac{5}{2}\left[2 \sqrt{x^{2}+4 x+10}\right]-7 \log \left|(x+2) \right.\)\(\left.\sqrt{x^{2}+4 x+10}\right|+C\)
\(\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=5 \sqrt{x^{2}+4 x+10}-7 \log \left|(x+2)\right.\)\(\left. \sqrt{x^{2}+4 x+10}\right|+C\)

\(\int \frac{d x}{x^{2}+2 x+2}=\int \frac{d x}{\left(x^{2}+2 x+1\right)+1}\)
\(\Rightarrow\int \frac{1}{(x+1)^{2}+(1)^{2}} d x\)
\(\Rightarrow\left[\tan ^{-1}(x+1)\right]+C\)

\(\int \frac{d x}{\sqrt{9 x}-4 x^{2}}=\int \frac{d x}{\sqrt{-4\left(x^{2}-\frac{9}{4} x\right)}}\)
\(\Rightarrow\int \frac{d x}{\sqrt{-4\left(x^{2}-\frac{9}{4} x+\frac{81}{64}-\frac{81}{64}\right)}}\)
\(\Rightarrow\int \frac{d x}{\sqrt{-4\left[\left(x-\frac{9}{8}\right)^{2}-\left(\frac{9}{8}\right)^{2}\right]}}\)
\(\Rightarrow\frac{1}{2}\left[\sin ^{-1}\left(\frac{x-\frac{9}{8}}{\frac{9}{8}}\right)\right]+C\)
\(\Rightarrow\frac{1}{2}\left[\sin ^{-1}\left(\frac{8 x-9}{9}\right)\right]+C\)