Evaluate $\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$ How to evaluate
$$\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$$
My attempt:
\begin{align*}
\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx &= \int_0^1 \frac{\ln \left(\dfrac{1-x^{n+1}}{1-x} \right)}{x}dx \\ 
&= \int_0^1 \frac{\ln \left({1-x^{n+1}} \right)}{x}dx -\int_0^1 \frac{\ln \left({1-x} \right)}{x}dx \\  
&=\frac{1}{n+1} \int_0^1 \frac{\ln \left({1-x} \right)}{x}dx -\int_0^1 \frac{\ln \left({1-x} \right)}{x}dx 
\end{align*}
but what's next?I have been stuck here for a while.

$$\int_0^1 \dfrac{\ln(1-x)}xdx = \int_0^1 \left(-\sum_{k=1}^{\infty} \dfrac{x^{k-1}}k\right)dx = -\sum_{k=1}^{\infty} \dfrac1k\int_0^1 x^{k-1}dx = - \sum_{k=1}^{\infty} \dfrac1{k^2} = -\zeta(2)$$
