Question

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

• A. $\frac{{\pi }^2}{32}$
• B. $\frac{\pi }{16}$
• C. $\frac{{\pi }^2}{16}$
• D. $\frac{\pi }{32}$

Answer 1

Answer: (A)

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

We have,

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

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

$dt=\frac{dx}{1+x^2}$

Therefore,

$I={\int }_0^{\frac{\pi }{4}}tdt$

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

$I=\frac{1}{2}({\left(\frac{\pi }{4}\right)}^2-0)$

$I=\frac{1}{2}\times \frac{{\pi }^2}{16}$

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

Hence, this is the answer.