Calculate $\int _0^\infty \frac{\ln x}{(x^2+1)^2}dx$ Calculate $$\int _0^\infty \dfrac{\ln x}{(x^2+1)^2}dx.$$
I am having trouble using Jordan's lemma for this kind of integral. Moreover, can I multiply it by half and evaluate $\frac{1}{2}\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$?

\begin{align}
I&=\int _0^\infty \dfrac{\ln x}{(x^2+1)^2}dx\\
&=\int _0^1 \dfrac{\ln x}{(x^2+1)^2}dx+\int _1^\infty \dfrac{\ln x}{(x^2+1)^2}dx\\
&=\int _0^1 \dfrac{\ln x}{(x^2+1)^2}dx-\int _0^1 \dfrac{x^2\ln x}{(x^2+1)^2}dx\\
&=\int _0^1 \dfrac{(1-x^2)\ln x}{(x^2+1)^2}dx\\
&=\int _0^1 \sum_{j=0}^{\infty}(-1)^j (2 j+1) x^{2 j}\ln xdx\\
&=\sum_{j=0}^{\infty}(-1)^j (2 j+1)\int _0^1  x^{2 j}\ln xdx\\
&=-\sum_{j=0}^{\infty}(-1)^j \frac{2 j+1}{(2j+1)^2}\\
&=-\sum_{j=0}^{\infty}(-1)^j \frac{1}{2j+1}\\
&=-\frac{\pi}{4}
\end{align}
where at last you may recall the expansion for $\arctan$.
