# How do you use partial fractions to find the integral int 3/(x^2+x-2) dx?

Jan 28, 2017

The integral equals $\ln | x - 1 | - \ln | x + 2 | + C$

#### Explanation:

Note that ${x}^{2} + x - 2$ can be factored as $\left(x + 2\right) \left(x - 1\right)$.

Thus , our partial fraction decomposition will be of the form

$\frac{A}{x + 2} + \frac{B}{x - 1} = \frac{3}{\left(x + 2\right) \left(x - 1\right)}$

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

$A x - A + B x + 2 B = 3$

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

Write a system of equations and solve:

$\left\{\begin{matrix}A + B = 0 \\ 2 B - A = 3\end{matrix}\right.$

Solve either through substitution or elimination to get $A = - 1$ and $B = 1$.

Therefore, the integral becomes

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

This can be solved using $\int \frac{1}{x} \mathrm{dx} = \ln | x | + C$.

$= \ln | x - 1 | - \ln | x + 2 | + C$

Hopefully this helps!