How to compute $\int_0^1\frac{t\ln t}{1+t^2}$ ? 
How to compute the integral $$\int_0^1\frac{t\ln t}{1+t^2}\ ?$$

So Wolfram alpha says it is exactly $-\dfrac{\pi^2}{48}$ .
I tried many substitutions without success, and partial integration as well.
Any help is welcome.
 A: I would expand $(1+t^2)^{-1}$ around $t=0$ and find out what $$\int_0^1 t^k \log t dt$$ is for each $k$.
ADD The sequence of functions $\displaystyle f_n=-t\log t\sum_{k=0}^n (-1)^k t^{2k}$ is dominated by $-t\log t$ over $[0,1]$ and converges to $f=-t\log t(1+t^2)^{-1}$ so Lebesgue's DCT ensures $$\int_0^1 \frac{t\log t}{1+t^2}dt= \sum_{k\geqslant 1}(-1)^k\int_0^1 t^{2k+1}\log t dt$$
A: Integrating by parts
$$
-\int_0^1\frac{t\ln t}{1+t^2}\,dt=\frac12\int_0^1\bigl(\log(1+t^2)\bigr)'\log t\,dt=\frac12\int_0^1\frac{\log(1+t^2)}{t}\,dt.
$$
Expad the integrand in a power series to get
$$
\int_0^1\frac{\log(1+t^2)}{t}\,dt=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_0^1t^{2n-1}\,dt=\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}.
$$
This is a well known sum.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#66f}{\large\int_{0}^{1}{t\ln\pars{t} \over 1 + t^{2}}\,\dd t}&=
\Re\int_{0}^{1}{\ln\pars{t} \over \ic + t}\,\dd t
=-\Re\int_{0}^{1}{\ln\pars{-\ic\bracks{\ic t}} \over 1 - \pars{\ic t}}
\,\pars{\ic\,\dd t}
=-\,\Re\int_{0}^{\ic}{\ln\pars{-\ic t} \over 1 - t}\,\dd t
\\[3mm]&=-\,\Re\int_{0}^{\ic}\ln\pars{1 - t}\,{1 \over -\ic t}\,\pars{-\ic}\,\dd t
=\Re\int_{0}^{\ic}{{\rm Li}_{1}\pars{t} \over t}\,\dd t
\end{align}
where $\ds{{\rm Li_{s}}\pars{z}}$ is a PolyLogarithm Function and
$\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$.

With the PolyLogaritm Recursive Property and $\ds{{\rm Li}_{2}\pars{0} = 0}$:
  \begin{align}
\color{#66f}{\large\int_{0}^{1}{t\ln\pars{t} \over 1 + t^{2}}\,\dd t}&=
\Re\int_{0}^{\ic}{\rm Li}_{2}'\pars{t}\,\dd t=\Re{\rm Li}_{2}\pars{\ic}
=\color{#66f}{\large -\,{\pi^{2} \over 48}}
\end{align}

Note that
\begin{align}&\color{#c00000}{\Re{\rm Li}_{2}\pars{\ic}}
=\Re\sum_{n = 1}^{\infty}{\ic^{n} \over n^{2}}
=\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \pars{2n}^{2}}
={1 \over 4}\bracks{\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}}
-\sum_{n = 1}^{\infty}{1 \over \pars{2n - 1}^{2}}}
\\[3mm]&={1 \over 4}\bracks{{1 \over 4}\sum_{n = 1}^{\infty}{1 \over n^{2}}
-\sum_{n = 1}^{\infty}{1 \over n^{2}}
+ \sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}}}
=-\,{1 \over 8}\sum_{n = 1}^{\infty}{1 \over n^{2}}
=-\,{1 \over 8}\,{\pi^{2} \over 6}=\color{#c00000}{-\,{\pi^{2} \over 48}}
\end{align}

Also, $\ds{\Re{\rm Li}_{2}\pars{\ic}}$ can be calculated by means of
  ${\sf\mbox{Jonquiere Inversion Formula}}$ as shown in the above cited link.
A 'straightforward' calculation can be performed by expanding
  $\ds{\ln\pars{1 - t}}$, in powers of $\ds{t}$, in the expression
  $\ds{-\,\Re\int_{0}^{\ic}{\ln\pars{1 - t} \over t}\,\dd t}$.

