Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$ I need to evaluate this integral: $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$.
I've tried $t=\log(x+1)$, $t=x+1$, but to no avail. I've noticed that:
$\int_0^1 \frac{\log(x+1)}{1+x^2} dx = \int_0^1\log(x+1) \arctan'(x)dx =\left. \log(x+1)\arctan(x) \right|_{x=0}^{x=1} - \int_0^1\frac{\arctan(x)}{x+1}dx$
But can't get further than this.
Any help is appreciated, thank you.

$\displaystyle A=\int_0^1\dfrac{\Big(\log(1+x)\Big)^2}{1+x^2}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$
$\displaystyle A=\int_0^1\dfrac{\left(\log\left(\dfrac{2}{1+x}\right)\right)^2}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2-\log(1+x)\Big)^2}{1+x^2}dx$
$\displaystyle A=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}dx-2\int_0^1\dfrac{\log 2\log(1+x)}{1+x^2}dx+A$
Therefore,
$\displaystyle \int_0^1\dfrac{2\log 2\log(1+x)}{1+x^2}dx=\int_0^1\dfrac{\Big(\log 2\Big)^2}{1+x^2}dx$
Finally,
$\displaystyle \int_0^1\dfrac{\log(1+x)}{1+x^2}dx=\int_0^1\dfrac{\log 2}{2(1+x^2)}dx=\dfrac{\log 2}{2}\Big[\arctan x\Big]_0^1=\dfrac{\pi\log 2}{8}$
