Question

Solve:

$\stackrel{1}{\underset{0}{\int }}\frac{\left({\tan }^{-1}x\right)}{1+x^2}dx$

• A. $1$
• B. $\frac{{\pi }^3}{64}$
• C. $\frac{{\pi }^2}{192}$
• D. $\frac{{\pi }^2}{32}$

Answer 1

The correct option is D $\frac{{\pi }^2}{32}$

${\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx$

$\Rightarrow {\tan }^{-1}x=t$

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

$\Rightarrow$ Limits: $t={\tan }^{-1}0$ to ${\tan }^{-1}1$

$\Rightarrow 0$ to $\frac{\pi }{4}$

${\int }_0^{-\pi /4}tdt={\left[\frac{t^2}{2}\right]}_0^{\pi /4}={\pi }^2/32$.