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.

Another way:
Integrate by parts
$$\int\ln x\cdot\dfrac x{(1+x^2)^2}\ dx=\ln x\int\dfrac x{(1+x^2)^2}\ dx-\int\left(\dfrac{d(\ln x)}{dx}\int\dfrac x{(1+x^2)^2}\ dx\right)dx$$
$$=-\dfrac{\ln x}{2(1+x^2)}+\int\dfrac{dx}{2x(1+x^2)}$$
Again $$\int\dfrac{dx}{x(1+x^2)}=\int\dfrac{(x^2+1-x^2)\ dx}{x(1+x^2)}=\int\dfrac{dx}x-\int\dfrac{x\ dx}{1+x^2}=\dfrac12\ln\dfrac{x^2}{1+x^2}$$
