Question

What is ${\int }_0^1\frac{{\tan }^{-1}x}{1+x^2}dx$ equal to ?

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

Answer 1

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

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

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

Now for $x=1,y=\frac{\pi }{4}$ and for $x=0,y=0$

Putting the values in the integral we have,

$I={\int }_0^{\frac{\pi }{4}}y.dy$

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

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

Hence, D is correct.