$$\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$$
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.
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} $$\sum_{k=0}^{\infty}(-1)^k(k+1)\left(- \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right) = -\dfrac1{1^2} + \dfrac{3}{3^2} - \dfrac{5}{5^2} + \dfrac7{7^2} \mp \cdots\\ = -1 + \dfrac13 - \dfrac15 + \dfrac17 \mp \cdots = -\dfrac{\pi}4$$
$$f(z)=\frac{\log z}{(1+z^2)^2} = \frac{\log z}{(z+i)^2(z-i)^2}$$
where we use the principal branch of the logarithm.
$$P.V.\int_{-R}^R f(z)\, dz = \int_{\epsilon}^R f(z)\, dz+\int_{-R}^\epsilon f(z)\, dz = \int_{\epsilon}^R f(z)\, dz + \int_{\epsilon}^R\frac{\log z+i\pi}{(1+z^2)^2}\, dz =$$
$$= 2\int_{\epsilon}^R f(z)\, dz + i\pi\int_{\epsilon}^R \frac{dz}{(1+z^2)^2}$$
Then, considering the contour $\Gamma$ that traverses $-R$ to $R$ (indented at $z=0$ around $\epsilon e^{i \theta}$) and then travels back along $Re^{i \theta}$, there is only one pole of $f(z)$ in $\Gamma$. As $R \to \infty$, the integral around the arc disappears, so
$$2\pi i \operatorname*{Res}_{z = i} f(z) = 2\int_{\epsilon}^\infty f(z)\, dz + \int_{0}^\pi f(\epsilon e^{i \theta})i\epsilon e^{i \theta}\, d\theta + i\pi\int_{\epsilon}^\infty \frac{dz}{(1+z^2)^2}$$
and as $\epsilon \to 0$, we see the RHS becomes
$$2\int_{0}^\infty f(z)\, dz + \frac{i \pi^2}{4}$$
The poles are of order 2, thus
$$\operatorname*{Res}_{z = i} f(z) = \lim_{z \to i}\,\frac{d}{dz}(z-i)^2f(z) = \frac{\pi+2 i}{8}$$
Thus
$$-\frac{\pi}{2}+ \frac{i \pi^2}{4} = 2\int_{0}^\infty f(z)\, dz + \frac{i \pi^2}{4}$$
and we finally find
$$\int_{0}^\infty \frac{\log z}{(1+z^2)^2}\, dz = -\frac{\pi}{4}$$
$\blacksquare$
Did you try taking $u=\frac{1}{x}$ ? You get usually something very similar to the first integral and then IBP could help.
Too late to the party (as I always am) but here's a solution that hasn't been posted yet.
Let $\displaystyle I(\lambda) = \int_{0}^{\infty} \frac{\ln{x}}{\lambda+x^2}\;{dx}$. Then $\displaystyle I'(\lambda) = -\int_{0}^{\infty} \frac{\ln{x}}{(\lambda+x^2)^2}\;{dx}$. Put $ x \mapsto \sqrt{\lambda} {x}$ then:
$$\begin{aligned}I(\lambda) & = \frac{1}{2\sqrt{\lambda}}\int_{0}^{ \infty}\frac{\ln(\lambda)}{1+x^2}\;{dx}-\frac{1}{\sqrt{\lambda}}\int_{0}^{ \infty}\frac{\ln{x}}{1+x^2}\;{dx} \\& = \frac{\pi}{4\sqrt{\lambda}}\log(\lambda)- \frac{1}{\sqrt{\lambda}}\int_{0}^{ \infty}\frac{\ln{x}}{1+x^2}\;{dx} \\& =\frac{\pi}{4\sqrt{\lambda}}\log(\lambda)-\frac{1}{\sqrt{\lambda}}\int_{0}^{\pi/2}\ln(\tan{x)} \\& = \frac{\pi}{4\sqrt{\lambda}}\log(\lambda)-\frac{1}{\sqrt{\lambda}}J \end{aligned}$$
Let $\displaystyle x \mapsto \frac{\pi}{2}-x$ then we have $\displaystyle J = \int_{0}^{\pi/2}\ln\left(\cot{x}\right)\;{dx}$ thus $\displaystyle 2J = \int_{0}^{\pi/2}\ln(1)\;{dx} = 0.$
Therefore $\displaystyle I'(\lambda) = \frac{\pi}{4\sqrt{\lambda}}\log(\lambda)$. We have $\displaystyle I'(\lambda) = \frac{\pi}{8 \lambda \sqrt{\lambda} }(2-\log{\lambda})$ and $\displaystyle -I'(1) = -\frac{\pi}{4}$.
P.S. $J$ can be calculated by $x \mapsto \frac{1}{x}$ as it gives us $J = -J$ readily but that substitution was embedded in previous solutions I wanted to present a fresh perspective. Also from the very beginning we could have let $ x \mapsto \sqrt{x}\tan{x}$. By differentiating both sides $n$-times we could get a nice generalisation too.
Let’s consider a simpler one $$ I(a)=\int_0^{\infty} \frac{\ln x}{a^2+x^2} d x $$ Putting $x=\frac{a}{y} $ transforms $I(a)$ into $$ \begin{aligned} I(a) & =\int_0^{\infty} \frac{\ln \left(\frac{a}{y}\right)}{a^2+\frac{a^2}{y^2}} \frac{a d y}{y^2} \\ & =\int_0^{\infty} \frac{\ln a-\ln y}{1+y^2} d y \\ & =\frac{\pi}{2} \ln a-\int_0^{\infty} \frac{\ln y}{1+y^2} d y \\ & =\frac{\pi}{2} \ln a \end{aligned} $$ Differentiating both sides w.r.t. $a$ gives $$ I^{\prime}(a)=-\int_0^{\infty} \frac{2 a \ln x}{\left(a^2+x^2\right)^2} d x=\frac{\pi}{2 a} $$ Re-arranging yields $$ \boxed{\int_0^{\infty} \frac{\ln x}{\left(a^2+x^2\right)^2}dx=-\frac{\pi}{4 a^2}} $$ In particular, $$\int_0^{\infty} \frac{\ln x}{\left(1+x^2\right)^2}dx=-\frac{\pi}{4 }$$
