<s>[INST] $\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$ Evaluate Integral Anyone remember the method for this? I think this should been done on the site
$$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$$
 [/INST] I think @kiwi ment
$$
\fbox {$I$} = \int_0^\infty \frac {\ln x}{x^2+a^2} dx = \left | u = \frac {a^2}x \Longrightarrow\left\{\begin{array}{c}
\ln x = 2 \ln a - \ln u \\
dx = -\frac {a^2du}{u^2}
\end{array}\right\} \right | = -\int_\infty^0 \frac{2\ln a - \ln u}{\frac {a^4}{u^2}+a^2}\frac {a^2 du}{u^2} = \\
2\int_0^\infty \frac{\ln a}{u^2+a^2}du-\int_0^\infty \frac{\ln u}{u^2+a^2}du = \fbox{$2\int_0^\infty \frac{\ln a}{u^2+a^2}du - I$}
$$
From last part it's clear that
$$
I = \int_0^\infty \frac{\ln a}{u^2+a^2}du
$$
This integral can be easily found
$$
I = \ln a\int_0^\infty \frac {du}{u^2+a^2} = \frac {\ln a}a \ \left.\mbox{atan}\ \frac ua \right|_0^\infty = \frac {\pi \ln a}{2a}
$$
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