How do you integrate #int (x)/(x^2-2x-3)# using partial fractions?
1 Answer
Feb 17, 2016
# 3/4 ln|x-3| + 1/4 ln|x + 1 | + c#
Explanation:
begin by factorising the denominator
#x^2 - 2x - 3 = (x - 3 )(x + 1 )#
Since the factors are linear , the numerators of the partial fractions will be constants , say A and B.
#rArr x/((x-3)(x+1)) = A/(x - 3 ) + B/(x+1) # multiply through by (x-3)(x+1)
x = A(x+1) + B(x-3) .................................(1)
The aim now is to find values of A and B. Note , that if x = - 1 the term with A will be zero and if x = 3 , the term with B will be zero. This is the starting point to finding A and B.
let x = - 1 in (1) : - 1 = - 4B
#rArr B = 1/4 # let x = 3 in (1) : 3 = 4A
# rArr A = 3/4 #
#rArr int(x/(x^2-2x-3) dx = int(3/4)/(x-3) dx + int(1/4)/(x+1) dx #
# = 3/4 ln|x-3| + 1/4 ln|x+1| + c #
