# How do you evaluate the integral int (x-1)/(x+1)dx?

Feb 6, 2017

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

#### Explanation:

Split the integrand function:

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

Using the linearity of the integral:

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

These are standard integrals that we can solve directly:

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