<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] \begin{align}
I&=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\ dx\overset{\large x\ \mapsto\frac1x}{=}-I\\
2I&=0\\
I&=0
\end{align}
</s>