# How do you integrate int ( x+10)/(x^2+2x-8) using partial fractions?

Feb 6, 2016

2ln|x-2| - ln|x+4| + c

#### Explanation:

the first step is to factor the denominator

${x}^{2} + 2 x - 8 = \left(x + 4\right) \left(x - 2\right)$

since these factors are linear then the numerator will be a constant

hence $\frac{x + 10}{\left(x + 4\right) \left(x - 2\right)} = \frac{A}{x + 4} + \frac{B}{x - 2}$

the next step is to multiply both sides by (x+4)(x-2)

x + 10 = A(x-2) + B(x+4)

Note that when x = -4 or x =2 the terms with A and B will be zero

let x = 2 : 12 = 6B → B = 2

let x = -4 : 6 = -6A → A = -1

$\Rightarrow \frac{x + 10}{{x}^{2} + 2 x - 8} = \frac{2}{x - 2} - \frac{1}{x + 4}$

Integral can be written as :

$\int \frac{2}{x - 2} \mathrm{dx} - \int \frac{\mathrm{dx}}{x + 4}$

= 2ln|x-2| - ln|x+4| + c