Compute $\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$ Given
$$\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$$
I couldn't evaluate this integral. My only idea here was evaluating this as integration by parts.
\begin{align}
\int\frac{x \log(x)}{(1+x^2)^2}dx & = \frac{1}{2} \int\frac{ \log(x)}{(x^2+1)^2}d(x^2+1)\\
& = \frac{1}{2} \frac{ \log(x)}{x^2+1} - \frac{1}{2}\int \left[(x^2+1)\frac{\frac{(x^2+1)^2}{x} - 4x(x^2+1) \log(x)}{(x^2+1)^4} \right ]dx\\
& = \frac{ \log(x)}{2(x^2+1)}-\frac{1}{2}\int \left [ \frac{x^2+1-4x^2 \log(x)}{x(x^2+1)^2} \right ] dx\\
& = \frac{ \log(x)}{2(x^2+1)} - \frac{1}{2}\int \frac{dx}{x(x^2+1)} + 2\int\frac{x \log(x)}{(x^2+1)^2}dx
\end{align}
Or this method doesn't work here, or I have done a mistake somewhere. However, I have also tried doing $u = x^2+1$ substitution, but this also didnt gave me any good results.Thank you.

Here is a general technique. Let's consider the integral

$$ I = \int_{0}^{\infty} \frac{x^{s-1}}{(1+x^2)^2} dx = -\frac{1}{4}\,{\frac { \left( s-2 \right) \pi }{\sin \left( \pi \,s/2
 \right) }},$$

which is the Mellin transform of $\frac{1}{(1+x^2)^2}$. Your integral can be evaluate using $I$ as

$$ \int_{0}^{\infty} \frac{x\ln x }{(1+x^2)^2} dx = \lim_{s\to 2}\frac{dI}{ds} = 0. $$

