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#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 40 Maths Textbook Solution.

Answer: $x\tan ^{-1}x-\frac{1}{2}\ln\left ( 1+x^{2} \right )+\frac{\left ( tan^{-1}x \right )^{2}}{2}+c$

Given: $\int \frac{x^{2}\tan ^{-1}x}{1+x^{2}}dx$

Hint: Let $1+x^{2}=u$ and $\tan ^{-1}x=t$

Solution:

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

Let $1+x^{2}=u\Rightarrow 2xdx=dx$

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

$\int \frac{\tan ^{-1}x}{1+x^{2}}dx$

Let,$\tan ^{-1}x=\Rightarrow dt=\frac{1}{1+x^{2}}dx$

\begin{aligned} &\int t d t=\frac{t^{2}}{2}=\frac{\left(\tan ^{-1} x\right)^{2}}{2} \\ &\Rightarrow \int x d x-\int \frac{x}{1+x^{2}} d x=x \tan ^{-1} x-\frac{1}{2} \ln \left(1+x^{2}\right)-\frac{\left(\tan ^{-1} x\right)^{2}}{2}+c \end{aligned}