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 $\frac{1}{2}\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$?

Here is a 'real-analysis route'.
Step 1. We have $$ \int_0^{+\infty}\frac{\ln x}{x^2+1} dx=0 \tag1$$ as may be seen by writing
$$
\begin{align}
 \int_0^{+\infty}\frac{\ln x}{x^2+1} dx&=\int_0^1\frac{\ln x}{x^2+1} dx+\int_1^{+\infty}\frac{\ln x}{x^2+1} dx\\\\
&=\int_0^1\frac{\ln x}{x^2+1} dx-\int_0^1\frac{\ln x}{\frac{1}{x^2}+1} \frac{dx}{x^2}\\\\
&=0.
\end{align}
$$
Step 2. Assume $a>0$. We have $$ \int_0^{+\infty}\frac{\ln x}{x^2+a^2} dx=\frac{\pi}{2}\frac{\ln a}{a} \tag2$$ 
as may be seen by writing
$$
\begin{align}
 \int_0^{+\infty}\frac{\ln x}{x^2+a^2} dx&= \int_0^{+\infty}\frac{\ln (a \:u)}{(a \:u)^2+a^2} (a \:du)\\\\
&=\frac1a\int_0^{+\infty}\frac{\ln a +\ln u}{u^2+1} du\\\\
&=\frac{\ln a}{a}\int_0^{+\infty}\frac{1}{u^2+1} du+\frac1{a}\int_0^{+\infty}\frac{\ln u}{u^2+1} du\\\\
&=\frac{\ln a}{a}\times \frac{\pi}2+\frac1{a}\times 0\\\\
&=\frac{\pi}{2}\frac{\ln a}{a}.
\end{align}
$$
Step 3. Assume $a>0$. We have $$ \int_0^{+\infty}\frac{\ln x}{(x^2+a^2)^2} dx=\frac{\pi}{4}\frac{\ln a-1}{a^3} \tag3$$ 
since sufficient conditions are fulfilled to differentiate both sides of $(2)$.
Putting $a:=1$ in $(3)$ gives

$$ \int_0^{+\infty}\frac{\ln x}{(x^2+1)^2} dx=\color{blue}{-\frac{\pi}{4}}.
$$

