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?

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.
