Analytical solution to $\int \frac{x^k}{1-x}dx$?

Question

Is there any simple way to get this integral? I'm also using Wolfram Alpha, but I get the solution below. I need specific answer not function that depends on a summation like a hypergeometric function. $$ \int \frac{x^k}{1-x}dx = \frac{x^{k + 1}\sideset{_2}{_1}F(1, k + 1, k + 2, x)}{k + 1} + C $$ The problem related to summation below: $$ \begin{align} S=&\sum_{n=1}^{\infty}\frac{x^n}{k+n}\quad,k\in[0,2\pi]\\ S=&\sum_{n=1}^{\infty}\frac{x^n}{k+n}\\ x^kS=&\sum_{n=1}^{\infty}\frac{x^{k+n}}{k+n}\\ \frac{d}{dx} (x^kS)=&\sum_{n=1}^{\infty}\frac{(k+n)x^{k+n-1}}{k+n}\quad,\text{if $|x|\lt 1 $}\\ =&\sum_{n=1}^{\infty}x^{k+n-1}=\frac{x^k}{1-x}\\ \int\frac{d}{dx} (x^kS)dx=&\int\frac{x^k}{1-x}dx \end{align} $$

Answer 1

$$ \int {\frac{{x^k }}{{1 - x}}dx} $$ Let, $1-x = t$. Then, $-dx = dt$. On substituting this, $$ \int\frac{(1-t)^k}{t}(-dt) = \int\frac{1}{t}\sum_{r}\binom{k}{r}(-t)^r(-dt) = -\ln t + \sum_{r}\binom{k}{r}\frac{(-t)^r}{r} . $$