How to compute
$\int_0^1\frac1{1-x^n}dx\;,$
where $n>0$?
How to compute
$\int_0^1\frac1{1-x^n}dx\;,$
where $n>0$?
Since $x^n \leq x$ in $[0,1]$, $\int^1_0 \frac{1}{1-x^n} dx \geq \int^1_0 \frac{1}{1-x} dx$ and the right integral is divergent. So your integral diverges.
Note that
$\frac 1 {1-x^n} = \frac 1 {n(1-x)} +f(x)$
for a function $f(x)$ that is continuous in the interval $[0,1]$, so the improper integral will exist for all $n \ge 1$ if and only if it exists for $n=1$.
But for $n=1$ you have an explicit antiderivative (using the logarithm), and you can check that it tends to infinity if you let the upper limit of the integral go to 1.