Question

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

Answer 1

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

Explanation:

the first step is to factor the denominator

`x^2 + 2x - 8 = (x+4)(x-2)`

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

hence `(x+10)/((x+4)(x-2)) = A/(x+4) + 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

`rArr (x+10)/(x^2+2x-8) = 2/(x-2) - 1/(x+4)`

Integral can be written as :

`int2/(x-2) dx -intdx/(x+4)`

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