#### explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 20 maths

Answer:$-\frac{1}{4} \frac{\left(1+2 x^{2}\right)}{\left(1+x^{2}\right)^{2}}+C$

Hint: Use substitution method to solve this integral.

Given: $\int \frac{x^{3}}{\left(x^{2}+1\right)^{3}} d x$

Solution:

\begin{aligned} &\text { Let } I=\int \frac{x^{3}}{\left(x^{2}+1\right)^{3}} d x \\ &\text { Put } 1+x^{2}=t \Rightarrow 2 x \; d x=d t \\ &\Rightarrow d x=\frac{d t}{2 x} \text { then } \end{aligned}

\begin{aligned} I &=\int \frac{x^{3}}{t^{3}} \frac{d t}{2 x}=\int \frac{x^{2}}{t^{3}} d t \\ &=\frac{1}{2} \int \frac{t-1}{t^{3}} d t \quad\left[\because 1+x^{2}=t \Rightarrow x^{2}=t-1\right] \end{aligned}

\begin{aligned} &=\frac{1}{2} \int\left\{\frac{t}{t^{3}}-\frac{1}{t^{3}}\right\} d t=\frac{1}{2} \int\left\{t^{1-3}-t^{-3}\right\} d t \\ &=\frac{1}{2}\left[\int\left(t^{-2}-t^{-3}\right) d t\right] \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\frac{t^{-2+1}}{-2+1}-\frac{t^{-3+1}}{-3+1}\right]+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{1}{2}\left[\frac{t^{-1}}{-1}-\frac{t^{-2}}{-2}\right]+\mathrm{c}=\frac{1}{2}\left[\frac{-1}{t}+\frac{1}{2 t^{2}}\right]+c \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\frac{-1}{1+x^{2}}+\frac{1}{2\left(1+x^{2}\right)^{2}}\right]+c \\ &=\frac{1}{2}\left[\frac{-2\left(1+x^{2}\right)+1}{2\left(1+x^{2}\right)^{2}}\right]+c \end{aligned}

\begin{aligned} &=\left[\frac{-2-2 x^{2}+1}{4\left(1+x^{2}\right)^{2}}\right]+c=\left[\frac{-1-2 x^{2}}{4\left(1+x^{2}\right)^{2}}\right]+c \\ &=-\frac{1}{4} \frac{\left(1+2 x^{2}\right)}{\left(1+x^{2}\right)^{2}}+c \end{aligned}