How do you find the indefinite integral of int (2x)/(x-1)^2?

1 Answer
Feb 1, 2017

$2 \ln | x - 1 | - \frac{2}{x - 1} + c$

Explanation:

Substitute $u = x - 1$. Then $\mathrm{dx} = \mathrm{du}$, $x = u + 1$:

$\int \frac{2 \left(u + 1\right)}{u} ^ 2 \mathrm{du}$
$= 2 \int \frac{1}{u} \mathrm{du} + 2 \int \frac{\mathrm{du}}{u} ^ 2$
$= 2 \ln | u | - \frac{2}{u} + c$
$= 2 \ln | x - 1 | - \frac{2}{x - 1} + c$