<s>[INST] Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substitution} \\
&= \int \tan t \cos^2 t \ln \tan t \,dt \\
&= \int \sin t \cos t \ln \tan t \,dt \\
&= -\frac{1}{2} \cos^2 t \ln \tan t + \frac{1}{2} \int \cot t \, dt
&& \text{by parts} \\
&= -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t + k
\end{align}
I run into a wall when I introduce the limits of the integral, since I get
\begin{align}
\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx
&= \bigg[ -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t \bigg] ^\frac{\pi}{2} _0
\end{align}
I'm not too sure how to evaluate the limit of the final equation as $t \rightarrow 0$. I feel like the solution is something very trivial, but I can't quite put my finger on what I'm forgetting.
 [/INST] Different approach:
\begin{align}
I&=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\ dx\overset{x^2=y}{=}\frac14\int_0^\infty\frac{\ln y}{(1+y)^2}\ dy\overset{1+y=z}{=}\frac14\int_1^\infty\frac{\ln(z-1)}{z^2}\ dz\\
&\overset{\large z\ \mapsto\ \frac{1}{z}}{=}\frac14\int_0^1\ln\left(\frac{1-z}{z}\right)\ dz=\frac14\underbrace{\int_0^1\ln(1-z)\ dz}_{\large1-z\ \mapsto\ z}-\frac14\int_0^1\ln z\ dz\\
&=\frac14\int_0^1\ln z\ dz-\frac14\int_0^1\ln z\ dz=0
\end{align}
</s>