Question

The value of integral ${\int }_0^{\infty }\frac{x\log x}{(1+x^2{)}^2}dx$ is

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

Answer 1

The correct option is A $0$

$I={\int }_0^{\infty }\frac{x\log xdx}{(1+x^2{)}^2}$
Let $x=\frac{1}{t}$
$I={\int }_0^{\infty }\frac{\left(\frac{1}{t}\right)\log \left(\frac{1}{t}\right)(-\frac{1}{t^2})dt}{(1+(\frac{1}{t^2}){)}^2}$
$I=-{\int }_0^{\infty }\frac{t\log t}{(1+t^2{)}^2}dt=-I$
or $I=0$