Evaluating $\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$ $$\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$$
I've done a similar one: $\int^{\infty}_{1}{\frac{\ln^n x}{x^2}dx}$ using IBP, but in this case I've tried IBP multiple times, differentiating all of the possible choices, and it's only getting more convoluted. Also, no substitution I've tried like $x=e^u$ or $u=1+x^2$made the integral simpler. I'd appreciate a hint at this point, since I don't think it can be that hard. 

Let $x=1/y$. We then get
$$I = \int_0^{\infty} \dfrac{\ln(x)}{(1+x^2)^2} dx = \int_0^{1} \dfrac{\ln(x)}{(1+x^2)^2} dx +  \int_1^{\infty} \dfrac{\ln(x)}{(1+x^2)^2} dx\\
 = \int_0^{1} \dfrac{\ln(x)}{(1+x^2)^2} dx -  \underbrace{\int_0^{1} \dfrac{x^2\ln(x)}{(1+x^2)^2} dx}_{x \to 1/x}\\
= \int_0^{1} (1-x^2)\left(\sum_{k=0}^{\infty} (-1)^{k}(k+1)x^{2k}\right) \log(x) dx\\
= \sum_{k=0}^{\infty}(-1)^k(k+1)\left(\int_0^1 x^{2k} \log(x) dx - \int_0^1 x^{2k+2} \log(x) dx \right)\\
= \sum_{k=0}^{\infty}(-1)^k(k+1)\left(- \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right) = -\dfrac{\pi}4$$
The last equality can be seen from below.
$$\int_0^1 x^m \log(x) dx = -\dfrac1{(m+1)^2}$$
\begin{align}
\sum_{k=0}^{\infty}(-1)^k(k+1)\left(- \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right) & = 1\left(-\dfrac1{1^2} + \dfrac1{3^2} \right)\\
& -2 \left(-\dfrac1{3^2} + \dfrac1{5^2} \right)\\
& +3 \left(-\dfrac1{5^2} + \dfrac1{7^2} \right)\\
& -4 \left(-\dfrac1{7^2} + \dfrac1{9^2} \right) + \cdots
\end{align}
$$\sum_{k=0}^{\infty}(-1)^k(k+1)\left(- \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right) = -\dfrac1{1^2} + \dfrac{3}{3^2} - \dfrac{5}{5^2} + \dfrac7{7^2} \mp \cdots\\
= -1 + \dfrac13 - \dfrac15 + \dfrac17 \mp \cdots = -\dfrac{\pi}4$$
