Evaluating $\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$

Substituting $x\leftarrow \tfrac{1}{x}$ shows that $$\int_0^\infty \frac{\log(x)}{(1+x^2)^2}dx = -\int_0^\infty \frac{x^2\log(x)}{(1+x^2)^2}dx$$ and therefore $$2\int_0^\infty \frac{\log(x)}{(1+x^2)^2}dx = \int_0^\infty \frac{1-x^2}{(1+x^2)^2}\log(x)dx.$$ Partial integration of the right hand side using that $$\frac{\partial}{\partial x}\left(\frac{x}{1+x^2}\right) = \frac{1-x^2}{(1+x^2)^2}$$ results in $$\int_0^\infty \frac{1-x^2}{(1+x^2)^2}\log(x)dx = \left[\frac{x\log(x)}{1+x^2}\right]_0^\infty -\int_0^\infty \frac{dx}{1+x^2} = 0 -\frac{\pi}{2}.$$ So the requested integral equals $-\frac{\pi}{4}$.

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} $$\sum_{k=0}^{\infty}(-1)^k(k+1)\left(- \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right) = -\dfrac1{1^2} + \dfrac{3}{3^2} - \dfrac{5}{5^2} + \dfrac7{7^2} \mp \cdots\\ = -1 + \dfrac13 - \dfrac15 + \dfrac17 \mp \cdots = -\dfrac{\pi}4$$