<s>[INST] 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.
 [/INST] 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
$$I = \displaystyle \int_0^1 \dfrac{x\log(x)}{(1+x^2)^2}dx = -\dfrac{\log(2)}4$$
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