#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 30 maths.

Answer: $\frac{\pi ^2}{32}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given: $\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}dx$

Solution: $I=\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}dx$

Put $\tan^{-1}x=t$

$\Rightarrow \frac{1}{1+x^{2}}dx=dt$

$\Rightarrow dx=\left (1+x \right )^2dt$

When x=0  then t=0  and when x=1  and $t=\frac{\pi}{4}$

\begin{aligned} I &=\int_{0}^{\frac{\pi}{4}} \frac{t}{1+x^{2}}\left(1+x^{2}\right) d t \\ =& \int_{0}^{\frac{\pi}{4}} t d t \\ =&\left[\frac{t^{1+1}}{1+1}\right]_{0}^{\frac{\pi}{4}} \\ =&\left[\frac{t^{2}}{2}\right]_{0}^{\frac{\pi}{4}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[t^{2}\right]_{0}^{\frac{\pi}{4}} \\ &=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^{2}-0^{2}\right] \\ &=\frac{1}{2}\left(\frac{\pi^{2}}{16}\right) \\ &=\frac{\pi^{2}}{32} \end{aligned}