#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 7

Answer : -       $\frac{1}{\sqrt{2}} \log \left|\frac{x+2-\sqrt{2(x+1)}}{x+2+\sqrt{2(x+1)}}\right|+C$

Hint :-                Use substitution and special integration formula.

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

Sol : -

\begin{aligned} &\text { Let } \mathrm{I}=\int \frac{x}{\left(x^{2}+2 x+2\right) \sqrt{x+1}} d x \\ &\text { Put }, x+1=\mathrm{t}^{2} \\ &\mathrm{dx}=2 \mathrm{t} \mathrm{dt} \text { then } \end{aligned}

\begin{aligned} &\mathrm{I}=\int \frac{t^{2}-1}{\left\{\left(t^{2}-1\right)^{2}+2\left(t^{2}-1\right)+2\right\} \sqrt{t^{2}}} .2 \text { tdt }\left(\because t^{2}-1=x\right) \\ &=\int \frac{\left(t^{2}-1\right)}{\left(t^{4}+1-2 t^{2}+2 t^{2}-2+2\right) t} \cdot 2 \mathrm{t} \mathrm{dt} \\ &=2 \int \frac{t^{2}-1}{\left(t^{4}+1\right)} \mathrm{dt} \end{aligned}

$=2 \int \frac{\frac{t^{2}-1}{t^{2}}}{\frac{t^{4}+1}{t^{2}}} \mathrm{dt}$    [Dividing num. And denom. By t2]

\begin{aligned} &=2 \int \frac{t^{\frac{2}{2}}-\frac{1}{t^{2}}}{\frac{t^{4}}{t^{2}}+\frac{1}{t^{2}}} \mathrm{dt} \\ &=2 \int \frac{\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}} \mathrm{dt} \end{aligned}

\begin{aligned} &=2 \int \frac{\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+2-2} d t \\ &=2 \int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left(t^{2}+\frac{1}{t^{2}}+2 . t_{t}^{1}\right)-2} d t \\ &=2 \int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left(t+\frac{1}{t}\right)^{2}-2} \text { dt }\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right] \end{aligned}

\begin{aligned} &\text { Again, put } \mathrm{t}+\frac{1}{t}=\mathrm{u} \\ &\Rightarrow\left(1-\frac{1}{t^{2}}\right) d t=\mathrm{du} \text { then } \\ &\qquad \mathrm{I}=2 \int \frac{d u}{u^{2}-(\sqrt{2})^{2}} \end{aligned}

\begin{aligned} &=2 \cdot \frac{1}{2 \cdot \sqrt{2}} \log \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+c \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+c\right]\\ &=\frac{1}{\cdot \sqrt{2}} \log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+c \quad\left[\because \mathrm{u}=\mathrm{t}+\frac{1}{t}\right]\\ &=\frac{1}{\sqrt{2}} \log \left|\frac{\frac{t^{2}+1-\sqrt{2} t}{t}}{\frac{t^{2}+1+\sqrt{2} t}{t}}\right|+c \end{aligned}

\begin{aligned} &=\frac{1}{\sqrt{2}} \log \left|\frac{t^{2}+1-\sqrt{2} t}{t^{2}+1+\sqrt{2} t}\right|+\mathrm{c} \\ &=\frac{1}{\cdot \sqrt{2}} \log \left|\frac{x+1+1-\sqrt{2} \cdot \sqrt{x+1}}{x+1+1+\sqrt{2} \cdot \sqrt{x+1}}\right|+\mathrm{c} \quad\left[\because \mathrm{t}=\sqrt{x+1} \text { or } t^{2}=x+1\right] \\ &=\frac{1}{\sqrt{2}} \log \left|\frac{x+2-\cdot \sqrt{2(x+1)}}{x+2+\sqrt{2(x+1)}}\right|+\mathrm{c} \end{aligned}