Evaluating an indefinite integral using complex analysis Using tools from complex analysis, I have to prove that 
$$ \int_0^{\infty} \frac{\ln x}{(x^2 + 1)^2}\,dx = - \frac{\pi}{4}.$$
But I'm not really sure where I should start. Any help would be appreciated.
 A: Let $C$ be the classical key-hole contour and $I$ be the integral
$$\begin{align}I&=\oint_C \frac{\log^2(z)}{(z^2+1)^2}\,dz\\\\
&=\int_0^R \frac{\log^2(x)}{(x^2+1)^2}\,dx+\int_R^0 \frac{(\log(x)+i2\pi)^2}{(x^2+1)^2}\,dx+\int_0^{2\pi}\frac{\log^2(Re^{i\phi})}{(R^2e^{i2\phi}+1)^2}iRe^{i\phi}\,d\phi\\\\
&=-i4\pi\int_0^R\frac{\log(x)}{(x^2+1)^2}\,dx+\int_0^R\frac{4\pi^2}{(x^2+1)^2}\,dx+\int_0^{2\pi}\frac{\log^2(Re^{i\phi})}{(R^2e^{i2\phi}+1)^2}iRe^{i\phi}\,d\phi \tag 1
\end{align}$$
As $R\to \infty$ the first integral on the right-hand side of $(1)$ approaches the integral of interest, the second integral approaches $\pi^3$, and the third approaches $0$.  
In addition, from the residue theorem, we have
$$\begin{align}
I&=2\pi i \text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=\pm i \right)\\\\
&=2\pi i\left(\left(-\frac{\pi}{4}+i\frac{\pi^2}{16}\right)+\left(\frac{3\pi}{4}-i\frac{9\pi^2}{16}\right)\right)\\\\
&=\pi^3+i\pi^2 \tag 2
\end{align}$$
Setting $(1)$ equal to $(2)$, we find
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty\frac{\log(x)}{(x^2+1)^2}\,dx=-\frac{\pi}{4}}$$
as was to be shown!
