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

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 $1/u\cdot\mathbb du=(n+1)/x\cdot\mathbb dx$ $$\int_0^1\frac{\ln(1-x^{n+1})}{x}\mathbb dx=\frac{1}{(n+1)}\int_0^1\frac{\ln(1-u)}{u}\mathbb du=-\frac{I}{(n+1)}. $$

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: $$\begin{align}\int_0^1-\frac{\ln(1-x)}{x}dx&=\int_0^11+\frac{x}{2}+\frac{x^2}{3}+\frac{x^4}{4}+\cdots dx\\ &=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\\ &=\frac{\pi^2}{6}\end{align}$$

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: $$ I = \left[ \sum_{k=1}^\infty \frac{x^{k(n+1)}}{(n+1)k^2} \right]_0^1 -\left[ \sum_{k=1}^\infty \frac{x^k}{k^2} \right]_0^1 $$ $$ I= \frac{\zeta(2)}{n+1} - \zeta(2) $$ $$ I = \left(\frac{-n}{n+1}\right) \frac{\pi^2}{6} $$