Question

The value of the integral $\underset{0}{\overset{1}{\int }}x{\cot }^{-1}(1-x^2+x^4)dx$ is :

• A. $\frac{\pi }{4}-\frac{1}{2}{\log }_{e}2$
• B. $\frac{\pi }{2}-{\log }_{e}2$
• C. $\frac{\pi }{4}-{\log }_{e}2$
• D. $\frac{\pi }{2}-\frac{1}{2}{\log }_{e}2$

Answer 1

The correct option is A $\frac{\pi }{4}-\frac{1}{2}{\log }_{e}2$

$I=\underset{0}{\overset{1}{\int }}x{\cot }^{-1}(1-x^2+x^4)dx$
$=\underset{0}{\overset{1}{\int }}x{\tan }^{-1}\left[\frac{1}{1-x^2+x^4}\right]dx$
$=\underset{0}{\overset{1}{\int }}x{\tan }^{-1}\left[\frac{1}{1-x^2(1-x^2)}\right]dx$
$=\underset{0}{\overset{1}{\int }}x{\tan }^{-1}\left[\frac{x^2-(x^2-1)}{1+x^2(x^2-1)}\right]dx$
$=\underset{0}{\overset{1}{\int }}x[{\tan }^{-1}{x}^2-{\tan }^{-1}(x^2-1\left)\right]dx$

Let $x^2=t\Rightarrow 2xdx=dt$

$\therefore I=\frac{1}{2}\left[\underset{0}{\overset{1}{\int }}\{{\tan }^{-1}t-{\tan }^{-1}(t-1\left)\right\}dt\right]$
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}{\tan }^{-1}tdt-\frac{1}{2}\underset{0}{\overset{1}{\int }}{\tan }^{-1}(-t)dt$
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}{\tan }^{-1}tdt+\frac{1}{2}\underset{0}{\overset{1}{\int }}{\tan }^{-1}tdt$
$=\underset{0}{\overset{1}{\int }}{\tan }^{-1}tdt$
$=[t{\tan }^{-1}t{]}_0^1-\underset{0}{\overset{1}{\int }}\frac{1}{1+t^2}\times tdt$
$=\frac{\pi }{4}-\frac{1}{2}[{\log }_e(1+t^2){]}_0^1$
$=\frac{\pi }{4}-\frac{1}{2}{\log }_{e}2$