I would like prove a result in integration

Question

I would like prove this result

$$\int_0^1 \frac{\left(\log (1+x)\right)^2}{x}\mathrm dx=\frac{\zeta(3)}{4}$$

Answer 1

note that \begin{align} \int \frac{\ln^{2}(1+x)}{x} \, dx = - 2 \, Li_{3}(1+x) + 2 \, Li_{2}(1+x) \, \ln(1+x) + \ln(-x) \, \ln^{2}(1+x) \end{align} for which \begin{align} I_{2} = \int_{0}^{1} \frac{\ln^{2}(1+x)}{x} \, dx = \frac{\zeta(3)}{4}. \end{align}

Answer 2

Another derivation is the following one: $$-\log(1-z)=\sum_{n=1}^{+\infty}\frac{z^n}{n},$$ $$-\frac{\log(1-z)}{1-z}=\sum_{n=1}^{+\infty}H_n z^n,$$ $$\frac{d}{dz}\log^2(1-z)=-2\frac{\log(1-z)}{1-z}=\sum_{n=1}^{+\infty}2H_n z^n,$$ $$\log^2(1-z)=\sum_{n=1}^{+\infty}\frac{2H_n}{n+1}z^{n+1},$$ $$I_2=\int_{0}^{1}\frac{\log^2(1+z)}{z}\,dz = \sum_{n=1}^{+\infty}\frac{2H_n(-1)^{n+1}}{(n+1)^2},$$ where the last series can be verified to be equal to $\frac{1}{4}\zeta(3)$ by following this precious link.