TITLE: How to compute$\int_{0}^{1}\dfrac{x\ln(x)}{(x^2+1)^2}dx$
QUESTION [7 upvotes]: How to compute
  $$\int_{0}^{1}\dfrac{x\ln(x)}{(x^2+1)^2}dx$$
  I'm interested in more ways of computing this integral.

My Thoughts
\begin{align}
\int_{0}^{1}\dfrac{x\ln(x)}{(x^2+1)^2}dx&=\int_{0}^{1} \:\ln \left(x\right)\frac{x}{\left(x^2+1\right)^2}dx\\
&=\int_{0}^{1} \:\ln \left(x\right)\frac{x}{\left(x^2+1\right)^2}dx\\
\mathrm{Apply\:Integration\:By\:Parts}:
\end{align}
$$\fbox{$u=\ln \left(x\right),\:\:u'=\frac{1}{x},\:\:v'=\frac{1}{\left(x^2+1\right)^2}x,\:\:v=-\frac{1}{2\left(x^2+1\right)}$ }$$
\begin{align}
\int_{0}^{1}\dfrac{x\ln(x)}{(x^2+1)^2}dx&
=\ln \left(x\right)\left(-\frac{1}{2\left(x^2+1\right)}\right)\biggl|_{0}^{1}-\int_{0}^{1} \frac{1}{x}\left(-\frac{1}{2\left(x^2+1\right)}\right)dx\\
&=-\frac{\ln \left(x\right)}{2\left(x^2+1\right)}\biggl|_{0}^{1}+\int_{0}^{1} \:\frac{1}{2x\left(x^2+1\right)}dx\\
&=-\frac{\ln \left(x\right)}{2\left(x^2+1\right)}\biggl|_{0}^{1}+\dfrac{1}{2}\int_{0}^{1} \:\frac{1}{x\left(x^2+1\right)}dx\\
\end{align}

now let's calculate:
\begin{align}
\int_{0}^{1} \frac{1}{x\left(x^2+1\right)}dx&=\int_{0}^{1} \frac{1}{x\left(x^2+1\right)}dx\\
&=\int_{0}^{1} \frac{1}{x}-\frac{x}{x^2+1}dx\\
&=\ln \left(x\right)\biggl|_{0}^{1}-\dfrac{1}{2}\int_{0}^{1} \frac{2x}{x^2+1}dx\\
&=\ln \left(x\right)\biggl|_{0}^{1}-\dfrac{1}{2}\int_{0}^{1} \frac{(x^2+1)'}{x^2+1}dx\\
&=\ln \left(x\right)\biggl|_{0}^{1}-\frac{1}{2}\ln \left(x^2+1\right)\biggl|_{0}^{1}\\
\int_{0}^{1} \frac{1}{x\left(x^2+1\right)}dx&=\left(\ln \left(x\right)-\frac{1}{2}\ln \left(x^2+1\right)\right)\biggl|_{0}^{1}\\
\end{align}
then $$\fbox{$\int_{0}^{1}\dfrac{x\ln(x)}{(x^2+1)^2}dx=-\frac{\ln \left(x\right)}{2\left(x^2+1\right)}\biggl|_{0}^{1}+\dfrac{1}{2}\left(\ln \left(x\right)-\frac{1}{2}\ln \left(x^2+1\right)\right)\biggl|_{0}^{1}$}$$

\begin{align}
\int_{0}^{1}\dfrac{x\ln(x)}{(x^2+1)^2}dx&=-\frac{\ln \left(x\right)}{2\left(x^2+1\right)}\biggl|_{0}^{1}+\dfrac{1}{2}\left(\ln \left(x\right)-\frac{1}{2}\ln \left(x^2+1\right)\right)\biggl|_{0}^{1} \\
&=\frac{1}{2}\left(\ln \left(x\right)-\frac{1}{2}\ln \left(x^2+1\right)\right)-\frac{\ln \left(x\right)}{2\left(x^2+1\right)}\biggl|_{0}^{1} \\
\end{align}
or the limit of 
\begin{align}
\lim _{x\to \:0+}\left(\frac{1}{2}\left(\ln \left(x\right)-\frac{1}{2}\ln \left(x^2+1\right)\right)-\frac{\ln \left(x\right)}{2\left(x^2+1\right)}\right)&=\lim _{x\to \:0+}\left(\frac{2x^2\ln \left(x\right)-x^2\ln \left(x^2+1\right)-\ln \left(x^2+1\right)}{4\left(x^2+1\right)}\right)\\
&=\dfrac{0}{4}=0
\end{align}
and
$$\frac{1}{2}\left(\ln \left(1\right)-\frac{1}{2}\ln \left(1^2+1\right)\right)-\frac{\ln \left(1\right)}{2\left(1^2+1\right)}=\dfrac{-\ln(2)}{4} $$
Finaly 
$$\fbox{$\int_{0}^{1}\dfrac{x\ln(x)}{(x^2+1)^2}dx=\dfrac{-\ln(2)}{4} $} $$

REPLY [2 votes]: The substitution $x=e^{-u}$ gives
$$\int_0^1{x\ln x\over(x^2+1)^2}dx=-{1\over4}\int_0^\infty u\,\text{sech}^2u\,du$$
Integration by parts tells us the indefinite integral is
$$\int u\,\text{sech}^2u\,du=u\tanh u-\int\tanh u\,du=u\tanh u-\ln\cosh u+C$$
so, on noting that $0\tanh0-\ln\cosh0=0$, it remains to evaluate the limit
$$\begin{align}
\lim_{u\to\infty}(u\tanh u-\ln\cosh u)&=\lim_{u\to\infty}\left(u{e^u-e^{-u}\over e^u+e^{-u}}-\ln\left(e^u+e^{-u}\over2 \right) \right)\\
&=\lim_{u\to\infty}\left(u\left(1-{2\over1+e^{2u}} \right)- u-\ln(1+e^{-2u})+\ln2\right)\\
&=\ln2
\end{align}$$
A side remark on the OP's solution:  Technically it's not correct to write something like
$$\int_{0}^{1}\dfrac{x\ln(x)}{(x^2+1)^2}dx=-\frac{\ln \left(x\right)}{2\left(x^2+1\right)}\biggl|_{0}^{1}+\dfrac{1}{2}\left(\ln \left(x\right)-\frac{1}{2}\ln \left(x^2+1\right)\right)\biggl|_{0}^{1}$$
because the two separate evaluations at the lower limit $x=0$ give a result of the form $\infty-\infty$.  The OP, of course, combines things here before evaluating.  But it's really better (in my opinion) not to write things down (or up) in a way that requires you to keep track of what the formulas "actually" mean.