How to taylor expand $1/(1- \frac{k}{r})$ where k is some constant Hi i am doing some integrations and there came across one step that i couldn't understand. The step is as follows:
$$\frac{dr}{1 - \displaystyle\frac{k}{r}}=dr\left(1 + \frac{k}{r-k}\right)$$. I don't know how we can come up with right hand side from left hand side. I know that using taylor series expansion, we have $\frac{1}{1-x}=1 +x+x^2 + x^3+...$ but i am not able to apply this here. How did we get the right hand side $dr(1 + \frac{k}{r-k})$ from left hand side of the same equation?
PS: k is a constant here and r is the variable term.

$$\frac{dr}{1 - \displaystyle\frac{k}{r}}=\frac{dr}{\frac{r-k}{r}}=\\dr(\frac{r}{r-k})=\\dr(\frac{r-k+k}{r-k})=\\
\\dr(\frac{r-k}{r-k}+\frac{k}{r-k})=\\
\\dr(1+\frac{k}{r-k})$$
