TITLE: Evaluating the integral $I (k)= \int_\infty^0\frac{k}{x^2 +k^2}\ln x ~ dx$
QUESTION [5 upvotes]: $$\int_0^\infty\left(\dfrac{\pi}{1 + \pi^2x^2}- \dfrac{1}{1+x^2}\right)\ln x dx$$

Attempt: 
$x \to \dfrac{1}{x}$
gives: 
$\displaystyle\int_\infty^0 \left(\dfrac{\pi}{\pi^2+x^2}- \dfrac{1}{1+x^2}\right)\ln x  ~dx$
So its basically reduced to evaluating: $I (k)= \displaystyle\int_\infty^0\dfrac{k}{x^2 +k^2}\ln x ~dx$
For this I tried: 

Differentiation wrt k  using Lebnitz Rule 
Integration by parts

Both these methods don't seem to work because of the presence of messy inverse functions. Please let me know how to continue.

REPLY [2 votes]: Here's a great way to proceed after substituting $x= kt$ : 
$I(k) = \displaystyle \int_0^\infty \dfrac{\ln(kt)}{1+t^2} dt$
$\implies \dfrac{dI}{dk}=\dfrac 1k \displaystyle\int_0^\infty\dfrac{1}{1+t^2}$dt
$\implies \dfrac{dI}{dk}= -\dfrac 1k\dfrac{\pi}{2}$
$$\implies  \bbox[5px,border:2px solid black]
{I(k)= -\dfrac{\pi}{2}\ln k }$$