Question

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

• A. $\frac{{\pi }^2}{32}$
• B. $\frac{\pi }{2}$
• C. $\frac{{\pi }^2}{16}$
• D. None of the above

Answer 1

Answer: (A)

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

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

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

Then, $d\left({\tan }^{-1}x\right)=dt\Rightarrow dx=(1+x^2)dt$

$x=0\Rightarrow t={\tan }^{-1}0=0$

$x=1\Rightarrow t={\tan }^{-1}1=\frac{\pi }{4}$

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