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.
 A: \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}
A: Set $\dfrac1x=y$
$$\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx=I=\int_\infty^0\dfrac{y^4\ln(y^{-1})}{y(1+y^2)^2}\left(\dfrac{dy}{-y^2}\right)=-\int_0^\infty\dfrac{y\ln y}{(1+y^2)^2}\ dy =-I$$
A: 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 $$\int\dfrac{dx}{x(1+x^2)}=\int\dfrac{(x^2+1-x^2)\ dx}{x(1+x^2)}=\int\dfrac{dx}x-\int\dfrac{x\ dx}{1+x^2}=\dfrac12\ln\dfrac{x^2}{1+x^2}$$
A: Using $x\mapsto\frac1x$  for $x>1$,$$\int_{0}^{\infty}\frac{x^{a}\ln^{b}x}{\left(1+x^{2}\right)^{c}}dx=\int_{0}^{1}\left(\frac{x^{a}\ln^{b}x}{\left(1+x^{2}\right)^{c}}+\left(-1\right)^{b}\frac{x^{2c-a-2}\ln^{b}x}{\left(1+x^{2}\right)^{c}}\right)dx$$vanishes provided $b$ is odd with $c=a+1$, as in your problem viz. $a=b=1,\,c=2$.
A: 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}
A: Others have already provided elegant and short solutions, but I thought that maybe that the user fysh would be interested in another proof showing a more general way of computing this type of integral. First, consider the following integral:
$$I(\alpha)\equiv\int_0^\infty \frac{x^\alpha}{(1+x^2)^2}\,dx.$$
We can notice that your integral, $I$, is just 
$$I=I'(1)$$
First, substitute $z=x^2$, leading to 
$$I(\alpha)=\frac{1}{2}\int_0^\infty\frac{z^{\alpha/2-1/2}}{(1+z)^2}\,dz.$$
Using a representation of the Beta function as well as its relationship to the Gamma function,  we find that
$$I(\alpha)=\frac{1}{2}B(\alpha/2+1/2, -\alpha/2+3/2)=\frac{1}{2}\Gamma(\alpha/2+1/2)\Gamma(3/2-\alpha/2)$$
At this point, we would like to use the reflection formula 
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)},$$
but are unable to with the present product. However, if we use the recursive property of the Gamma function, $x\Gamma(x)=\Gamma(1+x)$, or $\Gamma(x)=(x-1)\Gamma(x-1)$, we can turn our expression into
\begin{align}
\frac{1}{2}\Gamma(\alpha/2+1/2)\Gamma(3/2-\alpha/2)&=\frac{1}{2}\Gamma(\alpha/2+1/2)\Gamma(1/2-\alpha/2)(1/2-\alpha/2) \\
&=\left(\frac{1-\alpha}{4}\right)\Gamma(\alpha/2+1/2)\Gamma(1-(\alpha/2+1/2)) \\
&=\left(\frac{1-\alpha}{4}\right)\frac{\pi}{\sin(\pi\alpha/2+\pi/2)} \\
&=\left(\frac{1-\alpha}{4}\right)\frac{\pi}{\cos(\pi\alpha/2)}
\end{align}
Differentiating this last expression yields
$$I'(\alpha)=\frac{\pi}{4}\left[\frac{(1-\alpha)\sin(\pi\alpha/2)\pi/2-\cos(\pi\alpha/2)}{\cos^2(\pi\alpha/2)}\right]$$
After applying l'Hopital's rule in trying to take the $\alpha\to 1$ limit, the following limit is reached:
$$\lim_{\alpha\to 1}I'(\alpha)=-\frac{\pi}{8}\lim_{\alpha \to 1}\frac{(1-\alpha)\cos(\pi\alpha/2)}{\sin(\pi\alpha/2)\cos(\pi\alpha/2)}=0$$
Thus, 
\begin{align}
\int_0^\infty \frac{x\log(x)}{(1+x^2)^2}\,dx=0
\end{align}
Notice that this approach can be used to calculate any integral of the form
$$\int_0^\infty \frac{x^\alpha\log^m(x)}{(\beta^\gamma + x^\chi)^\mu}\,dx$$
with $\text{Re}\,\alpha>-1$, $\text{Re}\,\chi, \text{Re}\,\mu>0$, $\text{Re}\left(\chi\mu - \alpha\right)>1$, $m\in\mathbb{N}$ and $\beta, \gamma\in\mathbb{C}$. 
A: First consider substitution $t = 1 + x^2$ which gives $dt = 2x\,dx$ and thus
$$I:=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\,dx = \int_1^\infty \frac{\ln(t-1)^{1/2}}{t^2}\frac 12\,dt = \frac 14\int_1^\infty\frac{\ln(t-1)}{t^2}\, dt.$$
Now, notice that we can convert the last integrand to some rational function by partial integration, since derivative of $\ln(t-1)$ is $\frac 1{t-1}$, and antiderivative of $\frac 1{t^2}$ is $-\frac 1t$, so for $t>1$:
\begin{align}
\int\frac{\ln(t-1)}{t^2}\, dt &= -\frac{\ln(t-1)}{t} + \int\frac{dt}{t(t-1)} \\
&= -\frac{\ln(t-1)}{t} + \int\frac{dt}{t-1} - \int\frac{dt}t \\
&= -\frac{\ln(t-1)}{t} + \ln(t-1) - \ln t + C \\
&= \frac{(t-1)\ln(t-1)}{t}-\ln t + C
\end{align}
and thus $$4I = \lim_{t\to\infty}\left( \frac{(t-1)\ln(t-1)}{t}-\ln t \right) - \lim_{t\to 1^+} \left( \frac{(t-1)\ln(t-1)}{t}-\ln t \right).$$
Both limits evaluate to $0$:
$$\lim_{t\to 1^+} \left( \frac{(t-1)\ln(t-1)}{t}-\ln t \right) = \lim_{t\to 1^+} \frac{(t-1)\ln(t-1)}{t} = \lim_{t\to 1^+} \frac{\ln(t-1)}{1+\frac 1{t-1}} = \lim_{t\to 1^+} \frac{\frac 1{t-1}}{-\frac 1{(t-1)^2}} = 0,$$
and
$$\lim_{t\to \infty} \left( \frac{(t-1)\ln(t-1)}{t}-\ln t \right) = \lim_{t\to \infty} \frac{(t-1)\ln(t-1)-t\ln t}{t} = \lim_{t\to \infty} ((t-1)\ln(t-1)-t\ln t)' = \lim_{t\to\infty}((\ln(t-1)+1)-(\ln t+1)) = \lim_{t\to\infty}\ln\frac{t-1}{t} = \ln 1 = 0,$$
where we used L'Hôpital's rule in both limits. So, $I = 0$.
