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.
Best Answer
Going a little round-about way. Consider, for $ s \geqslant 0$, a parametric modification of the integral at hand: $$ \mathcal{I}(s) = \int_0^1 \frac{\log(1+s x)}{1+x^2} \mathrm{d} x $$ The goal is to determine $\mathcal{I}(1)$. Now: $$ \begin{eqnarray} \mathcal{I}(1) &=& \int_0^1 \mathcal{I}^\prime(s) \mathrm{d} s = \int_0^1 \left( \int_0^1 \frac{x}{1+s x} \frac{\mathrm{d} x}{1+x^2} \right) \mathrm{d} s \\ &=& \int_0^1 \left.\left( - \frac{1}{1+s^2} \log(1+s x) + \frac{s}{1+s^2} \arctan(x) + \frac{1}{2} \frac{\log(1+x^2)}{1+x^2} \right) \right|_{x=0}^{x=1} \mathrm{d} s \\ &=& \int_0^1 \left( \color\green{ -\frac{\log(1+s)}{1+s^2}} + \frac{1}{4} \frac{\pi s+\log(4)}{1+s^2}\right) \mathrm{d} s = - \mathcal{I}(1) + \frac{1}{4} \pi \log(2) \end{eqnarray} $$ Hence $$ \mathcal{I}(1) = \frac{\pi}{8} \log(2) $$