How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$ I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}$$
and
$$I_n=\int_0^1\frac{x-1}{x^{n+1}-1}dx$$
It makes sense to me that, for all values of $n$, $I_n$ is convergent since the bottom of the function is always above zero and the integral exists for $n\to\infty$ however I cannot seem to find a nice closed form for this.
One thought I did have was using:
$$\sum\ln(x_i)=\ln\left(\prod x_i\right)$$
but I cannot seem to make it work. Does anyone have any hints for this type of problem as I would like to try and complete it myself. Thanks :)

Decompose the integrand as
$$\frac{1}{1+x+x^2+...+x^n}=\sum_{k=1}^{n}\frac{a_k}{x-x_k},\>\>\>\>\> a_k=\frac{x_k(x_k-1)}{n+1}
$$
where $x_k= e^{i \frac{2\pi k}{n+1}},\> k=1,2...n$. Then
\begin{align}
I_n&=\int_0^1\frac{dx}{1+x+x^2+...+x^n}\\
&=\int_0^1 \sum_{k=1}^{n}\frac{a_k}{x-x_k}dx
=\frac1{n+1}\sum_{k=1}^{n} x_k(x_k-1)\ln\left(1-\frac1{x_k}\right)
\end{align}
Substitute $x_k= e^{i \frac{2\pi k}{n+1}}$ into above summation to derive the close-form
$$\color{blue}{I_n = \frac\pi{2(n+1)} \csc\frac{2\pi}{n+1}-\frac4{n+1}\sum_{k=1}^{[\frac n2]}
\sin\frac{\pi k}{n+1}\sin\frac{3\pi k}{n+1}\ln\left(\sin\frac{\pi k}{n+1}\right)}
$$
Listed below are some sample results
\begin{align}
I_2 &= \frac{\pi}{3\sqrt3}\\
I_3 &= \frac\pi8 +\frac14\ln2\\
I_4 &= \frac\pi{10}\csc\frac{2\pi}{5}+\frac1{\sqrt5}\ln\left(2\cos\frac\pi5\right)\\
 I_5 &= \frac{\pi}{6\sqrt3}+\frac13\ln2\\
 I_7 &= \frac{\pi}{8\sqrt2}+\frac18\ln2 + \frac{1}{8\sqrt2}\ln \frac{\sqrt2+1}{\sqrt2-1}\\
 I_9 &= \frac\pi{20}\csc\frac{\pi}{5}+\frac15\ln2
+\frac1{2\sqrt5}\ln\left(2\cos\frac\pi5\right)\\
\end{align}
\begin{align}
I_6
&=\frac\pi{14}\csc\frac{2\pi}{7}-\frac1{2\sqrt7}\bigg( \frac{\ln\sin\frac\pi7}{\sin\frac{2\pi}7}
 +\frac{\ln\sin\frac{2\pi}7}{\sin\frac{3\pi}7}-\frac{\ln\sin\frac{3\pi}7}{\sin\frac{\pi}7}\bigg)
\\
\end{align}
\begin{align}
I_8 &= \frac\pi{18}\csc\frac{2\pi}{9}+\frac{2}{3\sqrt3} \bigg(\frac{\ln\csc\frac\pi9}{\csc\frac{\pi}9}
 +\frac{\ln\csc\frac{2\pi}9}{\csc\frac{2\pi}9}-\frac{\ln\csc\frac{4\pi}9}{\csc\frac{4\pi}9}\bigg)
\\
\end{align}
