Prove that $\int_0^{+\infty} \frac{\ln x}{a^2+x^2} dx = \frac{\pi\ln a}{2a}$ Is that true $$\int_0^{+\infty} \cfrac{\ln x}{a^2+x^2} dx = \cfrac{\pi\ln a}{2a},$$ where $a>0$ ? And how to compute it?

Another approach :
Consider
$$
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx.\tag1
$$
Rewrite $(1)$ as
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac1{a^2}\int_0^\infty\frac{x^{b-1}}{1+\left(\frac{x}{a}\right)^2}\ dx.\tag2
\end{align}
Putting $x=ay\;\color{blue}{\Rightarrow}\;dx=a\ dy$ yields
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=a^{b-2}\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy,
\end{align}
where
\begin{align}
\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy=\frac{\pi}{2\sin\left(\frac{b\pi}{2}\right)}.
\end{align}
Hence
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac\pi2\cdot\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}.\tag3
\end{align}
Differentiating $(3)$ with respect to $b$ and setting $b=1$ yields
\begin{align}
\int_0^\infty\frac{\partial}{\partial b}\left[\frac{x^{b-1}}{a^2+x^2}\right]_{b=1}\ dx&=\frac\pi2\frac{\partial}{\partial b}\left[\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}\right]_{b=1}\\
\int_0^\infty\frac{\ln x}{x^2+a^2}\ dx&=\large\color{blue}{\frac{\pi\ln a}{2a}}.
\end{align}
