I've been working on this for the past hour.

$\displaystyle \int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx \ (a>0)$

let $\displaystyle u = \frac{a}{x} $

then $\displaystyle du = -\frac{a}{x^{2}}dx = -\frac{u^{2}}{a} dx $

$\displaystyle \int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx = -\int_{\infty}^{0} \frac{\ln\Big( \frac{a}{u}\Big)}{a^{2}+ \frac{a^{2}}{u^{2}}} \ \frac{a}{u^{2}} du = $ $\displaystyle \frac{1}{a} \int_{0}^{\infty} \frac{\ln\Big( \frac{a}{u}\Big)}{u^{2}+1} \ du $

$\displaystyle = \frac{\ln a}{a} \int_{0}^{\infty} \frac{1}{u^{2}+1} \ du -\frac{1}{a} \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du$

$\displaystyle \int_{0}^{\infty} \frac{1}{u^{2}+1} \ du = \tan^{-1} (\infty) - \tan^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

$\displaystyle \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du $

let $\displaystyle w = \frac{1}{u} $

then $\displaystyle dw = -\frac{1}{u^{2}} du = - w^{2} du $

$\displaystyle \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du = -\int_{\infty}^{0} \frac{\ln \Big( \frac{1}{w}\Big)}{\frac{1}{w^{2}}+1} \ \frac{1}{w^{2}} \ dw $ $\displaystyle = -\int_{0}^{\infty} \frac{\ln w}{w^{2}+1} \ dw $

which implies that $\displaystyle \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du =0 $

so finally $\displaystyle \int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx = \frac{\ln a}{a} \Big(\frac{\pi}{2}\Big) -\frac{1}{a} \Big(0\Big) = \frac{\pi \ln a}{2a} $ ?