Evaluate $\int_0^\infty\frac{\ln x}{1+x^2}dx$ Evaluate $$\int_0^\infty\frac{\ln x}{1+x^2}\ dx$$
I don't know where to start with this so either the full evaluation or any hints or pushes in the right direction would be appreciated.  Thanks.
Hint

Write $$\int_0^\infty\frac{\ln x}{(1+x^2)}dx=\int_0^1\frac{\ln x}{(1+x^2)}dx+\int_1^\infty\frac{\ln x}{(1+x^2)}dx$$ For the second integral make a change of variable $x=\frac{1}{y}$ and see the beauty of the result.

I am sure that you can take from here.
Another general 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&amp;=\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&amp;=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&amp;=\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&amp;=\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&amp;=\large\color{blue}{\frac{\pi\ln a}{2a}}.
\end{align}
Thus
$$
\int_0^\infty\frac{\ln x}{x^2+1}\ dx=\large\color{blue}{0}.
$$