Question

What is the value of \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\]?

पर्याय

• \[\frac{\pi}{8}\]

• \[\frac{1}{2}\]

• \[\frac{\pi^2}{32}\]

• \[\frac{\pi^2}{16}\]

Answer 1

The transformed integral is \[\left[\frac{t^2}{2}\right]_{0}^{\frac{\pi}{4}}\]. Thus its value is \[\frac{1}{2}\left(\frac{\pi}{4}\right)^2=\frac{\pi^2}{32}\].