<s>[INST] 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 $\frac{n \pi^2}{6(n+1)}$
 [/INST] Ok we have:
$$
I = \int_0^1 \frac{\ln(\sum_{k=0}^n x^n)}{x}dx
$$
Using:
$$
S_n = \frac{1-r^{n+1}}{1-r}
$$
$$
I = \int_0^1 \frac{\ln\left( \frac{1-x^{n+1}}{1-x} \right)}{x}dx
$$
Using the fact that $\ln(a/b) = \ln(a) - \ln(b)$
$$
I = \int_0^1 \frac{1}{x} \ln\left(1-x^{n+1} \right)dx - \int_0^1 \frac{1}{x} \ln(1-x) \, dx
$$
We now use $\ln$'s taylor series:
$$
\ln(1-x) = \sum_{k=1}^\infty \frac{x^k}{k}
$$
We obtain
$$
I = \int_0^1\sum_{k=1}^\infty \frac{x^{(n+1)k-1}}{k} dx - \int_0^1 \sum_{k=1}^\infty \frac{x^{k-1}}{k} dx
$$
Switching the bounds with the summation because of monotone convergence and integrating:
$$
I =  \left[ \sum_{k=1}^\infty \frac{x^{k(n+1)}}{(n+1)k^2} \right]_0^1 -\left[ \sum_{k=1}^\infty \frac{x^k}{k^2} \right]_0^1
$$
$$
I= \frac{\zeta(2)}{n+1} - \zeta(2)
$$
$$
I = \left(\frac{-n}{n+1}\right) \frac{\pi^2}{6}
$$
</s>