TITLE: Evaluating $\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$
QUESTION [11 upvotes]: $$\int^{\infty}_{0}{\frac{\ln x}{(1+x^2)^2}dx}$$
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.

REPLY [2 votes]: 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.