Question

${\int }_0^{\infty }\frac{x\ln x}{(1+x^2{)}^2}dx=$

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $0$

$1+x^2=t$
$2xdx=dt$
$xdx=\frac{dt}{2}$
$x=0,t=1$
$x=\infty ,t=\infty$
$x=\sqrt{t-1}$
$\ln x=\frac{1}{2}\ln (t-1)$
$\Rightarrow {\int }_1^{\infty }\frac{1}{2}\frac{\ln (t-1)}{t^2}\frac{dt}{2}$
$=\frac{1}{4}{\int }_1^{\infty }\frac{1}{t^2}\ln (t-1)dt$
Integrating by parts
$=\frac{1}{4}{[-\frac{1}{t}\ln (t-1)]}_1^{\infty }-\frac{1}{4}{\int }_1^{\infty }-\frac{1}{t(t-1)}dt$
$=\frac{1}{4}{[-\frac{1}{(1+x^2)}\ln \left(x^2\right)]}_0^{\infty }+\frac{1}{4}[{\int }_1^{\infty }\frac{1}{t-1}dt-{\int }_1^{\infty }\frac{1}{t-1}dt]$
$=\frac{1}{4}\times 0+\frac{1}{4}\times 0=0$