Question

The value of the integral $\underset{0}{\overset{\infty }{\int }}\frac{x\log x}{(1+x^2{)}^2}dx$ is

• A. $0$
• B. $\log 7$
• C. $5\log 13$
• D. $2$

Answer 1

The correct option is A $0$

$I=\underset{0}{\overset{\infty }{\int }}\frac{x\log x}{(1+x^2{)}^2}dx$
Put $x=1/y$
$\therefore dx=-\frac{1}{y^2}dy$
when $x\to 0,y\to \infty$ and when $x\to \infty ,y\to 0$
$\therefore I=\underset{\infty }{\overset{0}{\int }}\frac{\frac{1}{y}\log \frac{1}{y}}{{(1+\frac{1}{y^2})}^2}.\frac{-1}{y^2}dy=\underset{\infty }{\overset{0}{\int }}\frac{y\log y}{(1+y^2{)}^2}dy$
$=-\underset{0}{\overset{\infty }{\int }}\frac{y\log y}{(1+y^2{)}^2}dy$
$=-\underset{0}{\overset{\infty }{\int }}\frac{x\log xdx}{(1+x^2{)}^2}$
$\Rightarrow I=-I\therefore I=0$