Evaluate $\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$ 
How to evaluate:
$$\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$$


Attempt:
$$\int_{0}^{1}\frac{\ln(1 + x + x^2 + \ldots + x^n)}{x} \mathrm dx
= \int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\mathrm d x$$

Any hints would be appreciated.
Edit: Testing it with different values of $n$, it seems like the integral evaluates to be $\frac{n \pi^2}{6(n+1)}$

I derive under the integral
$$f(n)=\int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\,dx$$
$$f'(n)=\int_{0}^{1}\frac{\partial}{\partial n}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\,dx=$$
$$=\int_0^1\frac{x^n \log x}{x^{n+1}-1}\,dx=\frac{\pi ^2}{6 (n+1)^2}$$
$$f(n)=\int\frac{\pi ^2}{6 (n+1)^2}\,dn=C-\frac{\pi ^2}{6 (n+1)}$$
As $$f(1)=\int_0^1 \frac{\log (x)}{x-1} \, dx=\frac{\pi ^2}{6}$$
then $C=\frac{\pi ^2}{6}$ and finally we have
$$f(n)=\frac{\pi ^2}{6}-\frac{\pi ^2}{6 (n+1)}=\frac{\pi ^2 n}{6 (n+1)}$$
