Question #45e95

Mar 25, 2016

$\int {x}^{2} / \left(1 - x\right) \mathrm{dx} = - {x}^{2} / 2 - x - \ln | x - 1 | + C$

Explanation:

Using the property that $\int \left(f + g\right) = \int f + \int g$ along with the known integrals $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$ for $n \ne - 1$ and $\int \frac{1}{x} \mathrm{dx} = \ln | x | + C$, we have:

$\int {x}^{2} / \left(1 - x\right) \mathrm{dx} = \int \left(- x - 1 - \frac{1}{x - 1}\right) \mathrm{dx}$

$= - \int x \mathrm{dx} - \int 1 \mathrm{dx} - \int \frac{1}{x - 1} \mathrm{dx}$

$= - {x}^{2} / 2 - x - \ln | x - 1 | + C$