The value of definite integral $\int_0^1 \frac{ an ^{-1} x}{1+x^2} d x$ is

Question

The value of definite integral $\int_0^1 \frac{	an ^{-1} x}{1+x^2} d x$ is

Professor PC Thomas Classes · Answer

Substitute $u = 	an^{-1}x \Rightarrow du = \frac{1}{1+x^2}dx$. Integral becomes $\int_0^{\pi/4} u du = [\frac{u^2}{2}]_0^{\pi/4} = \frac{\pi^2}{32}$.