# How do you find the antiderivative of 1/(1-x)^2?

I found: $\frac{1}{1 - x} + c$
$\int \frac{1}{1 - x} ^ 2 \mathrm{dx} =$
Where: $d \left[1 - x\right] = - \mathrm{dx}$ giving:
$\int {\left(1 - x\right)}^{-} 2 \cdot - \mathrm{dx} = \frac{1}{1 - x} + c$