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 $\frac{n \pi^2}{6(n+1)}$
 A: A short supplement to A-Level-Student's answer:
The first integral
$$\int_0^1\frac{\ln(1-x^{n+1})}{x}\mathbb dx$$
can be represented by the second one
$$I:=-\int_0^1\frac{\ln(1-x)}{x}\mathbb dx$$
by the substitution $u:= x^{n+1}$ and thus $1/u\cdot\mathbb du=(n+1)/x\cdot\mathbb dx$
$$\int_0^1\frac{\ln(1-x^{n+1})}{x}\mathbb dx=\frac{1}{(n+1)}\int_0^1\frac{\ln(1-u)}{u}\mathbb du=-\frac{I}{(n+1)}.
$$
A: I don't know how to evaluate
$$\int_0^1\frac{\ln(1-x^{n+1})}{x}dx$$
(sorry) but I can do the other one:
$$\int_0^1-\frac{\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$
This result follows when you consider the Maclaurin expansion of the integrand and on recalling the solutions to the Basel problem, as shown below:
$$\begin{align}\int_0^1-\frac{\ln(1-x)}{x}dx&=\int_0^11+\frac{x}{2}+\frac{x^2}{3}+\frac{x^4}{4}+\cdots dx\\
&=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\\
&=\frac{\pi^2}{6}\end{align}$$
A: 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:
$$
I =  \left[ \sum_{k=1}^\infty \frac{x^{k(n+1)}}{(n+1)k^2} \right]_0^1 -\left[ \sum_{k=1}^\infty \frac{x^k}{k^2} \right]_0^1
$$
$$
I= \frac{\zeta(2)}{n+1} - \zeta(2)
$$
$$
I = \left(\frac{-n}{n+1}\right) \frac{\pi^2}{6}
$$
A: 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
$$f(n)=\frac{\pi ^2}{6}-\frac{\pi ^2}{6 (n+1)}=\frac{\pi ^2 n}{6 (n+1)}$$
