Integration of $\int_{0}^{1} \frac{x^k}{1-x} dx$ Solving this using integration by parts and by letting $u = x^k$ and $dv = (1-x)^{-1}$, it follows:
$$\int_{0}^{1}\frac{x^k}{1-x} dx = -x^{k}\ln(1-x) + k\int_{0}^{1} \ln(1-x)x^{k-1} dx$$
It is known that
$$\ln(1-x) = (1-x) - (1-x)\ln(1-x)$$
But doesn't this make the integration more complicated. Can I obtain a recursive formula for such an integration?
 A: Obtain the recursive relationship as follows
$$I_k= \int \frac{x^k}{1-x} dx= \int \frac{x^k-x^{k-1}}{1-x}dx + \int \frac{x^{k-1}}{1-x}dx\\
=-\int x^{k-1} dx + I_{k-1}
=-\frac1k{x^{k}}+ I_{k-1}\\
$$
A: Assuming that $k$ is an integer, you could write
$$\frac{x^k}{1-x}=\frac{x^k-1+1}{1-x}=\frac 1{1-x}-\frac{x^k-1}{x-1}=\frac 1{1-x}-\sum_{n=0}^{k-1}x^n$$
and integrate termwise.
A: Hint :
That is related to the Incomplete Beta function
A: Let $$ \mathcal{I} = \int_0^1 \frac{x^k}{1-x} \, \mathrm{d}x $$  Now using integration by parts we get
\begin{align*} 
 \mathcal{I}  &= -x^k\, \ln(1-x) \bigg{|}_0^1 + \int_0^1 k \, x^{k-1} \, \ln(1-x) \, \mathrm{d}x \\
&= 1 + \int_0^1 k \, x^{k-1} \left( \sum_{n=1}^\infty - \frac{x^n}{n} \right) \, \mathrm{d}x \\
&= 1 - \sum_{n=1}^\infty \frac{k}{n} \int_0^1 x^{k+n-1} \, \mathrm{d}x \\
&= 1 - \sum_{n=1}^\infty \frac{k}{n} \cdot \frac{1}{k+n} \\
&= 1 - \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{k+n} \right) \hspace{5mm} (1)
\end{align*}
Now remember Harmonic Series
(here I am using i  and m because we have used n and k in our integral)
$$ \mathrm{H}_m = \sum_{i=1}^{m} \frac{1}{i} $$
We can write $\mathrm{H}_m$ in infinite series representation, see
$$ \mathrm{H}_m = \sum_{i=1}^{\infty} \left( \frac{1}{i} - \frac{1}{i+m} \right) $$
Using this we can clearly see our (1) is
$$ \boxed{ \mathcal{I} = 1 - \mathrm{H}_k} $$
