Bazı noktaları atlayıp:
$\dfrac{1}{z+1}=\displaystyle\sum_{k=0}^{\infty}(-z)^k=\sum_{k=0}^{\infty}(-1)^kz^k$ ve
$\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^2}=\dfrac{\pi^2}{6}$ (Basel Problemi)
Kısmi İntegrasyon kullanarak:
$$\begin{aligned}\int z^k\ln z\,dz&=\dfrac{z^{k+1}}{k+1}\ln z-\int \dfrac{z^k}{k+1}\,dz=\dfrac{z^{k+1}}{k+1}\ln z-\dfrac{z^{k+1}}{(k+1)^2}+C\\&\Rightarrow \int_0^1 z^k\ln z\,dz=\dfrac{-1}{(k+1)^2}\end{aligned}$$
\[\begin{aligned}\int_0^1 \dfrac{\ln z}{z+1}\,dz&=\sum_{k=0}^{\infty}\dfrac{(-1)^{k+1}}{(k+1)^2}=\sum_{k=1}^{\infty}\dfrac{(-1)^{k}}{k^2}
\\&=-\left(\sum_{k=1}^{\infty}\dfrac{1}{k^2} \right)+2\left( \sum_{k=1}^{\infty}\dfrac{1}{(2k)^2} \right) \\&=-\sum_{k=1}^{\infty}\dfrac{1}{k^2}+\dfrac{1}{2} \sum_{k=1}^{\infty}\dfrac{1}{k^2}\\&=-\dfrac{\pi^2}{6}+\frac12\dfrac{\pi^2}{6}=-\dfrac{\pi^2}{12}\end{aligned}\]