Question

Solve : $\underset{0}{\stackrel{\infty }{\int }}\frac{x\ln x}{(1+x^2{)}^2}$

Answer 1

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

Integration by parts

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

$=-\frac{\ln x}{2(1+x^2)}-(-\frac{1}{2}(\left(\ln |x|\right)-\frac{1}{2}(\ln |x^2+1|)))$

$\therefore \int \frac{x\ln x}{(1+x^2{)}^2}dx=-\frac{\ln x}{2(1+x^2)}+\frac{1}{2}(\left(\ln |x|\right)-\frac{1}{2}(\ln |x^2+1|))+C$

$\therefore {\int }_0^{\infty }\frac{x\ln x}{(1+x^2{)}^2}dx=0-0=0$