A closed form for $\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx$? I would like some help to find a closed form for the following integral:$$\int_0^1 \frac{\left(\log (1+x)\right)^3}{x}dx $$ I was told it could be calculated in a closed form. I've already proved that $$\int_0^1 \frac{\log (1+x)}{x}dx  = \frac{\pi^2}{12}$$ using power series expansion. 

Thank you.
Letting $u = \log(1+x)$,

$$ \begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &amp;= \int_{0}^{\log 2} \frac{u^{3}}{e^{u}-1} e^{u} \ du \\ &amp;= \int_{0}^{\log 2} \frac{u^{3}}{1-e^{-u}} \ du \\ &amp;= \int_{0}^{\log 2} u^{3} \sum_{n=0}^{\infty} e^{-nu} \ du \\ &amp;= \sum_{n=0}^{\infty} \int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &amp;= \int_{0}^{\log 2} u^{3} \ du + \sum_{n=1}^{\infty}\int_{0}^{\log 2} u^{3} e^{-nu} \ du \\ &amp;= \frac{\log^{4}(2)}{4} + \sum_{n=1}^{\infty}\int_{0}^{\log 2} u^{3} e^{-nu} \ du .\end{align}$$

Then integrating by parts 3 times, 

$$\begin{align} \int_{0}^{1} \frac{\log^{3}(1+x)}{x} \ dx &amp;= \frac{\log^{4}(2)}{4} -\sum_{n=1}^{\infty} e^{-nu} \left(\frac{6}{n^{4}} + \frac{6u}{n^{3}} + \frac{3 u^{2}}{n^{2}} + \frac{u^{3}}{n} \right)\Bigg|^{\log 2}_{0} \\ &amp;= \frac{\log^{4}(2)}{4} - \sum_{n=1}^{\infty} \left[\frac{1}{2^{n}} \left(\frac{6}{n^{4}} + \frac{6 \log 2}{n^{3}} + \frac{3 \log^{2} (2)}{n^{2}} + \frac{\log^{3}(2)}{n}\right)  - 6 \zeta(4) \right] \\ &amp;= -\frac{3\log^{4}(2)}{4 } - 6 \text{Li}_{4} \left(\frac{1}{2} \right) - 6 \log (2) \ \text{Li}_{3} \left(\frac{1}{2} \right)-3 \log^{2}(2) \text{Li}_{2} \left(\frac{1}{2} \right) + 6 \zeta(4) \\ &amp;\approx 0.1425141979 . \end{align}$$

The answer could of course be simplified using the known values of $\text{Li}_{2} \left(\frac{1}{2} \right)$, $\text{Li}_{3} \left( \frac{1}{2}\right) $, and $\zeta(4)$.
do you know this results ?

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