Prove that $\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx = \frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$ I have to prove that
$$
I=\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}
$$
I know that 
$$ I_{+}=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{8}\log(2)
$$
and I can write
$$
I= \int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx+\int_{0}^{1}\frac{\log(1-x)}{1+x^2}dx
$$
So I have only to evaluate $$I_{-}=\int_{0}^{1}\frac{\log(1-x)}{1+x^2}dx
$$ 
Using the expansion of the logarithm valid in the range of integration I can also write 
$$I_{-}=-\sum_{n\geq1}\frac{1}{n}\int_{0}^{1}\frac{x^n}{1+x^2}dx
$$
but then I can go any further because I can't find a simple solution that depend only on $n$ for the rational (and solvable) integral under the sum sign .
Maybe is simple to considerate instead of $ I_{-} $ the integral $\int_{0}^{1}\frac{\log(1-x^2)}{1+x^2}dx=I$ because it seems to me that $\int_{0}^{1}\frac{x^{2n}}{1+x^2}dx= \sum_{k=0}^{n-1} (\frac{(-1)^{n+k}}{2k+1} + \frac{\pi}{4}(-1)^n)$ but again I can go further.
Any help would be appreciated.
Thanks 

\begin{align}
I&=\int_{-1}^1\frac{\ln(1+x)}{1+x^2}\ dx\overset{x=\frac{1-y}{1+y}}{=}\int_0^\infty\frac{\ln2-\ln(1+y)}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-\int_0^\infty\frac{\ln(1+y)}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-\left(\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy+\underbrace{\int_1^\infty\frac{\ln(1+y)}{1+y^2}\ dy}_{\large y\mapsto 1/y}\right)\\
&=\frac{\pi}{2}\ln2-\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy-\int_0^1\frac{\ln(1+y)-\ln y}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-2\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy+\int_0^1\frac{\ln y}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-2\left(\frac{\pi}{8}\ln2\right)-G\\
&=\frac{\pi}{4}\ln2-G
\end{align}
.

\begin{align}
\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&\overset{y=\frac{1-u}{1+u}}{=}\int_0^1\frac{\ln2-\ln(1+u)}{1+u^2}\ du\\
&=\frac{\pi}{4}\ln2-\int_0^1\frac{\ln(1+u)}{1+u^2}\ du\\
2\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&=\frac{\pi}{4}\ln2\\
\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy&=\frac{\pi}{8}\ln2
\end{align}
