#### need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 67

Answer: $\log \left|\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right|+c$

Hint: Use substitution method to solve this integral

Given: $\int \frac{1}{(x+1)\left(x^{2}+2 x+2\right)} d x$

Solution:

$\operatorname{let} I=\int \frac{1}{(x+1)\left(x^{2}+2 x+2\right)} d x$

\begin{aligned} &=\int \frac{1}{(x+1)\left(x^{2}+2 x+1+1\right)} d x \\ &=\int \frac{1}{(x+1)\left\{\left(x^{2}+2 x+1\right)+1\right\}} d x \end{aligned}

$=\int \frac{1}{(x+1)\left\{(x+1)^{2}+1\right\}} d x$                   $\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$\text { Putting } x+1=\tan u$        $......(i)$

$\Rightarrow d x=\sec ^{2} u \; d u \text { then }$

$I=\int \frac{1}{\tan u\left\{\tan ^{2} u+1\right\}} \sec ^{2} u\; d u$

$=\int \frac{1}{\tan u\left\{\sec ^{2} u\right\}} \sec ^{2} u \; d u\left[\because 1+\tan ^{2} u=\sec ^{2} u\right]$

$=\int \frac{1}{\tan u} d u=\int \cot u\; d u$

$=\log |\sin u|+c$            $........(ii)$     $\left[\because \int \cot x \; d x=\log |\sin x|+c\right]$

$\text { Also, from (i) } x+1=\tan u=\frac{\sin u}{\cos u} \quad\left[\because \tan x=\frac{\sin x}{\cos x}\right]$

\begin{aligned} &\Rightarrow(x+1) \cos u=\sin u \\ &\Rightarrow(x+1)^{2} \cos ^{2} u=\sin ^{2} u \quad[\text { Squaring on both sides }] \\ &\Rightarrow(x+1)^{2}\left(1-\sin ^{2} u\right)=\sin ^{2} u \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right] \\ &\Rightarrow(x+1)^{2}-(x+1)^{2} \sin ^{2} u=\sin ^{2} u \\ &\Rightarrow(x+1)^{2}=\sin ^{2} u+(x+1)^{2} \sin ^{2} u \\ &\Rightarrow(x+1)^{2}=\left[1+(x+1)^{2}\right] \sin ^{2} u \end{aligned}

$\Rightarrow \sin ^{2} u=\frac{(x+1)^{2}}{1+(x+1)^{2}}=\frac{(x+1)^{2}}{x^{2}+2 x+1+1}=\frac{(x+1)^{2}}{x^{2}+2 x+2} \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$\Rightarrow \sin u=\sqrt{\frac{(x+1)^{2}}{x^{2}+2 x+2}}=\frac{(x+1)}{\sqrt{x^{2}+2 x+2}}$    $........(iii)$

\begin{aligned} &\text { From (ii) and (iii) we get }\\ &I=\log \left|\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right|+c \end{aligned}