Find the value of $\int_{0}^{1} \frac{x \log x \:dx}{1+x^2}$ Find the value of $$I=\int_{0}^{1} \frac{x \log x \:dx}{1+x^2}$$
My Try: I used Integration by parts
So
$$I=\frac{1}{2}\log x \times \log (1+x^2) \biggr\rvert_{0}^{1}-\int_{0}^{1}\frac{\log(1+x^2)}{2x}dx$$
So
$$I=\frac{-1}{2}\int_{0}^{1}\frac{\log(1+x^2)}{x}$$
Can we proceed from here?
 A: Since $\int_{0}^{1}x^n \log(x)\,dx = -\frac{1}{(n+1)^2}$ and $\frac{x}{1+x^2}=x-x^3+x^5-x^7+\ldots$ on $(0,1)$, by termwise integration we have
$$ \int_{0}^{1}\frac{x\log x}{1+x^2}\,dx = -\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{6^2}+\frac{1}{8^2}-\ldots = \color{blue}{-\frac{\pi^2}{48}}$$
as a consequene of the Basel problem. The exchange of $\sum$ and $\int$ is allowed by the dominated convergence Theorem.
A: $\begin{align}I=\int_{0}^{1} \frac{x \log x }{1+x^2}\:dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}I=\dfrac{1}{4}\int_{0}^{1} \frac{\log x }{1+x}\:dx\end{align}$
And it is well known that,
$\begin{align}J=\int_{0}^{1} \frac{\log x }{1+x}\:dx=-\frac{\pi^2}{12}\end{align}$
Addendum:
Actually,
$\displaystyle K=\int_0^1 \frac{\ln x}{1-x}\,dx=-\zeta(2)$
(Taylor series expansion)
$\begin{align}K-J=\int_0^1 \dfrac{2x\ln x}{1-x^2}\,dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}K-J&=\dfrac{1}{2}\int_0^1 \dfrac{\ln x}{1-x}\,dx\\
&=\dfrac{1}{2}K
\end{align}$
Therefore,
$J=\dfrac{1}{2}K=-\dfrac{1}{2}\zeta(2)$
NB:
One assumes that $\displaystyle \zeta(2)=\dfrac{\pi^2}{6}$
