# What is int -x/ (x^2 -2x +1)dx?

Mar 19, 2018

$\int - \frac{x}{{x}^{2} - 2 x + 1} \mathrm{dx} = \frac{1}{x - 1} + \ln \left\mid \frac{1}{x - 1} \right\mid + \text{c}$

#### Explanation:

In order to find $\int - \frac{x}{{x}^{2} - 2 x + 1} \mathrm{dx}$, we must perform a partial fraction decomposition on the integrand.

$\frac{x}{{x}^{2} - 2 x + 1} = \frac{x}{x - 1} ^ 2 = \frac{\left(x - 1\right) + 1}{x - 1} ^ 2 =$

$\frac{x - 1}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 2 = \frac{1}{x - 1} + \frac{1}{x - 1} ^ 2$

So

$\int - \frac{x}{{x}^{2} - 2 x + 1} \mathrm{dx} = - \int \frac{1}{x - 1} ^ 2 + \frac{1}{x - 1} \mathrm{dx}$

$= \frac{1}{x - 1} - \ln \left\mid x - 1 \right\mid + \text{c}$

$= \frac{1}{x - 1} + \ln \left\mid \frac{1}{x - 1} \right\mid + \text{c}$