How do you integrate #int ( x+10)/(x^2+2x-8)# using partial fractions?
1 Answer
Feb 6, 2016
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
