# How do you integrate int (5x-1)/(x^2-x-2) using partial fractions?

May 13, 2016

$2 \ln \left(x + 1\right) + 3 \ln \left(x - 2\right) + c$, with $x > 2$

#### Explanation:

Let $f \left(x\right) = \frac{5 x - 1}{{x}^{2} - x - 2}$

$= \frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)}$

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

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

Determine matching A and B from $A \left(x - 2\right) + B \left(x + 1\right) = 5 x - 1$

Comparing coefficients of x, A+B=5. Comparing constants, $- 2 A + B = - 1$.

Solving, A = 2 and B = 3.

Now, $\int f \left(x\right) \mathrm{dx}$

$= 2 \int \frac{1}{x + 1} \mathrm{dx} + 3 \int \frac{1}{x - 2} \mathrm{dx}$

$= 2 \ln \left(x + 1\right) + 3 \ln \left(x - 2\right) + C$

For both the logarithms to be real, $x > 2$