Question

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

• A. $\frac{{\pi }^2}{4}$
• B. $\frac{{\pi }^2}{18}$
• C. $\frac{{\pi }^2}{32}$
• D. $\frac{{\pi }^2}{8}-1$

Answer 1

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

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

Let $z={\tan }^{-1}x$

For $x=0,z={\tan }^{-1}0=0$ and for $x=1,z={\tan }^{-1}1=\frac{\pi }{4}$

$⟹dz=\frac{1}{1+x^2}dx$

Hence, integration becomes-

${\int }_0^{\frac{\pi }{4}}zdz$

$={\left[\frac{z^2}{2}\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{2}[\frac{{\pi }^2}{16}-0]$

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