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.

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 $\log(1+t)$, 
$$
I(0) = -\frac{1}{4} \int_{0}^{1} \frac{1}{t}\left(t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots\right)dt = -\frac{1}{4}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} = -\frac{1}{4}\left(\sum_{n=1}^{\infty}\frac{1}{n^{2}}-2\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}}\right) = -\frac{1}{4}\cdot \frac{1}{2}\zeta(2) = -\frac{\pi^{2}}{48}.
$$
