Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$
Thanks.

Using twice the result that $\int_0^{\infty} \frac{\ln x}{1+x^2} d x=0$, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln x}{\left(1+x^2\right)^2} d x&=\int_0^{\infty} \frac{\left(1+x^2-x^2\right) \ln x}{\left(1+x^2\right)^2} d x \\
&=\int_0^{\infty} \frac{\ln x}{1+x^2} d x-\int_0^{\infty} \frac{x^2 \ln x}{\left(1+x^2\right)^2} d x \\
&=\frac{1}{2} \int_0^{\infty} x \ln x d\left(\frac{1}{1+x^2}\right) \\
&=\frac{1}{2} \int_0^{\infty} \frac{1-\ln x}{1+x^2} d x \\
&=\frac{1}{2} \int_0^{\infty} \frac{d x}{1+x^2} \\
&=\frac{\pi}{4}
\end{aligned}
$$
