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.
 A: 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

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

A: We have
$$\int_1^{\infty} \dfrac{x\log(x)}{(1+x^2)^2}dx = \int_1^0 \dfrac{1/x\log(1/x)}{(1+1/x^2)^2}\dfrac{-dx}{x^2} = \int_1^0 \dfrac{x\log(x)}{(1+x^2)^2} = -\int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx $$
Hence,
$$\int_0^{\infty} \dfrac{x\log(x)}{(1+x^2)^2}dx = \int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx + \int_1^{\infty} \dfrac{x\log(x)}{(1+x^2)^2}dx = 0$$

In case you are interested in $I = \displaystyle \int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx$, we have
\begin{align}
I & = \int_0^1 \dfrac12\dfrac{\log(x^2)}{(1+x^2)^2} d\left(\dfrac{x^2}2\right)
\end{align}
This gives us
\begin{align}
4I & = \int_0^1 \dfrac{\log(t)}{(1+t)^2} dt = \int_0^1 \sum_{k=0}^{\infty} (-1)^k(k+1)t^k\log(t) dt = \sum_{k=0}^{\infty}(-1)^k(k+1) \int_0^1 t^k \log(t)dt\\
& = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}(k+1)}{(k+1)^2} = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{k+1} = -\log(2)
\end{align}
This gives us
$$I = \displaystyle \int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx = -\dfrac{\log(2)}4$$
A: You can evaluate using the residue theorem.  In this case, by considering the contour integral
$$\oint_C dz \frac{z \log^2{z}}{(1+z^2)^2} $$
where $C$ is a keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$, and letting $R \to \infty$ and $\epsilon \to 0$, we have
$$-i 4 \pi I_1 + 4 \pi^2 I_0 = i 2 \pi \left [\frac{d}{dz} \left (\frac{z \log^2{z}}{(z+i)^2} \right ) \right ]_{z=e^{i \pi/2}} + i 2 \pi \left [\frac{d}{dz} \left (\frac{z \log^2{z}}{(z-i)^2} \right ) \right ]_{z=e^{i 3 \pi/2}} = 2 \pi^2$$
where
$$I_j = \int_0^{\infty} dx \frac{x \log^j{x}}{(1+x^2)^2} $$
Now, for similar reasons,
$$-i 2 \pi I_0 = i 2 \pi \left [\frac{d}{dz} \left (\frac{z \log{z}}{(z+i)^2} \right ) \right ]_{z=e^{i \pi/2}} + i 2 \pi \left [\frac{d}{dz} \left (\frac{z \log{z}}{(z-i)^2} \right ) \right ]_{z=e^{i 3 \pi/2}} = -i \pi$$
or $4 \pi^2 I_0 = 2 \pi^2$.  
Therefore
$$I_1 = \int_0^{\infty} dx \frac{x \log{x}}{(1+x^2)^2} = 0$$
