Question

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

• A. $\frac{{\pi }^2}{8}$
• B. $\frac{{\pi }^2}{16}$
• C. $\frac{{\pi }^2}{4}$
• D. $\frac{{\pi }^2}{32}$

Answer 1

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

$Putt=ta{n}^{-1}x\Rightarrow dt=\frac{1}{1+x^2}dx,then$
${\int }_0^1\frac{ta{n}^{-1}x}{1+x^2}dx={\int }_0^{\frac{\pi }{4}}tdt={\left[\frac{t^2}{2}\right]}_0^{\frac{\pi }{4}}=\frac{{\pi }^2}{32}$