Question

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

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

Answer 1

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

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

Put ${\tan }^{-1}x=p\Rightarrow \frac{dx}{1+x^2}=dp$

When $x=0,p=0$ and $x=1,p=\frac{\pi }{4}$

$\Rightarrow I={\int }_0^{\frac{\pi }{4}}pdp=\frac{p^2}{2}=\frac{{\pi }^2}{32}$

Hence, B is correct.