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 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: $$ I = \left[ \sum_{k=1}^\infty \frac{x^{k(n+1)}}{(n+1)k^2} \right]_0^1 -\left[ \sum_{k=1}^\infty \frac{x^k}{k^2} \right]_0^1 $$ $$ I= \frac{\zeta(2)}{n+1} - \zeta(2) $$ $$ I = \left(\frac{-n}{n+1}\right) \frac{\pi^2}{6} $$
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 $$f(n)=\frac{\pi ^2}{6}-\frac{\pi ^2}{6 (n+1)}=\frac{\pi ^2 n}{6 (n+1)}$$
$$ \begin{aligned} \int_0^1 \frac{\ln \left(1+x+x^2+\ldots+x^n\right)}{x} d x = & \int_0^1 \frac{\ln \left(1-x^{n+1}\right)-\ln (1-x)}{x} d x \\ = & \frac{1}{n+1} \int_0^1 \frac{\ln \left(1-x^{n+1}\right)}{x^{n+1}} d\left(x^{n+1}\right)-\int_0^1 \frac{\ln (1-x)}{x} d \\ = & \left(\frac{1}{n+1}-1\right) \int_0^1 \frac{\ln (1-x)}{x} d x \\ = & \frac{n}{n+1} \operatorname{Li_2}(1)=\frac{n \pi^2}{6(n+1)} \end{aligned} $$
