Calculating $\int^1_0\frac{x \log x}{1 + x^2} dx$ Trying to reinvent the spirit of Calculus, I was trying to Understand the Leibniz Rule of Differentiation under Integration sign. To check my understanding, I was considering this integral:
$$ I(b) = \int^1_0 \frac{x \log(b + x)}{1 + x^2} dx $$
After some simplification, I came upon the Integral in question:
$$ 2I(1) = \frac{1}{4} \log^2{2} + \frac{\pi^2}{16} + 2\int^1_0\frac{b \log b}{1 + b^2} db $$
Now I am a bit confused about how to progress. Maxima shows the answer to be $-\frac{\pi^2}{48}$.
This is not a homework problem.
 A: This calculation may contains some convergence problem, but it'll be resolved easily. 
Using integration by parts, we have 
$$
I(0)=\int_{0}^{1}\frac{x\log x}{1+x^{2}}dx = \left[\frac{1}{2}\log(1+x^{2})\log x\right]_{0}^{1} - \int_{0}^{1} \frac{1}{2}\log(1+x^{2})\frac{dx}{x}
$$
Since
$$
\lim_{x\to 0} \log(1+x^{2})\log x = \lim_{x\to 0} \frac{\log(1+x^{2})}{x^{2}}\cdot x\cdot x\log x = 0, 
$$
we have
$$
I(0) = -\frac{1}{2}\int_{0}^{1} \frac{\log(1+x^{2})}{x}dx. 
$$
Now use substitution $x^{2} = t$, then 
$$
I(0) = -\frac{1}{4} \int_{0}^{1}\frac{\log(1+t)}{t}dt
$$
Now use the Taylor series of $\log(1+t)$, 
$$
I(0) = -\frac{1}{4} \int_{0}^{1} \frac{1}{t}\left(t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots\right)dt = -\frac{1}{4}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} = -\frac{1}{4}\left(\sum_{n=1}^{\infty}\frac{1}{n^{2}}-2\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}}\right) = -\frac{1}{4}\cdot \frac{1}{2}\zeta(2) = -\frac{\pi^{2}}{48}.
$$
A: $$
\begin{align}
\int_0^1\frac{x\log(x)}{1+x^2}\,\mathrm{d}x
&=\sum_{k=0}^\infty\int_0^1(-1)^kx^{2k+1}\log(x)\,\mathrm{d}x\tag1\\
&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+2)^2}\tag2\\
&=\frac14\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\tag3\\
&=-\frac14\left(\sum_{k=1}^\infty\frac1{k^2}-2\sum_{k=1}^\infty\frac1{(2k)^2}\right)\tag4\\
&=-\frac{\pi^2}{48}\tag5
\end{align}
$$
Explanation:
$(1)$: expand $\frac{x}{1+x^2}$ into a geometric series
$(2)$: integrate by parts: $u=\log(x)$, $\mathrm{d}v=x^{2k+1}\,\mathrm{d}x$
$(3)$: substitute $k\mapsto k-1$
$(4)$: an alternating sum is all the terms minus twice the even terms
$(5)$: $\zeta(2)=\frac{\pi^2}6$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{0}^{1}{x\ln\pars{x} \over 1 + x^{2}}\,\dd x &
\,\,\,\stackrel{x\ \to\ x^{\large 1/2}}{=}\,\,\,
{1 \over 4}\int_{0}^{1}{\ln\pars{x} \over 1 + x}\,\dd x 
\,\,\,\stackrel{x\ \to\ -x}{=}\,\,\,
-\,{1 \over 4}\int_{0}^{-1}{\ln\pars{-x} \over 1 - x}\,\dd x
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
{1 \over 4}\int_{0}^{-1}{-\ln\pars{1 - x} \over x}\,\dd x =
{1 \over 4}\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\,\dd x =
{1 \over 4}\,\mrm{Li}_{2}\pars{-1}
\\[5mm] & =
{1 \over 4}\,\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}} =
{1 \over 4}\,\sum_{n = 1}^{\infty}\braces{%
{1 \over \pars{2n}^{2}} - \bracks{{1 \over n^{2}} - {1 \over \pars{2n}^{2}}}}
\\[5mm] & =
-\,{1 \over 8}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}
_{\ds{\pi^{2} \over 6}}\ =\
\bbx{-\,{\pi^{2} \over 48}}
\end{align}
