# How do you integrate int 3/(x^2+x-2) using partial fractions?

Jun 29, 2016

$= \ln \setminus \frac{x - 1}{x + 2} + C$

#### Explanation:

$\int \mathrm{dx} q \quad \frac{3}{{x}^{2} + x - 2}$

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

so we can say that

$\frac{3}{{x}^{2} + x - 2} = \frac{A}{x + 2} + \frac{B}{x - 1}$

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

$\setminus \implies A \left(x - 1\right) + B \left(x + 2\right) = 3$

let x = 1 so B*3 = 3, so B = 1

let x = -2 so A*(-3) = 3 so A = -1

so the integral becomes

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

$= \ln \left(x - 1\right) - \ln \left(x + 2\right) + C$

$= \ln \setminus \frac{x - 1}{x + 2} + C$