How do you integrate #int 3/ (x^2-6x+8)# using partial fractions?
1 Answer
Mar 11, 2016
Explanation:
Begin by factorising the denominator.
#x^2 - 6x + 8 = (x - 2 )(x - 4 )# Since these factors are linear then the numerators of the partial fractions will be constants , say A and B.
#rArr 3/((x-2)(x-4)) = A/(x-2) + B/(x-4)# Multiply through by (x - 2 )(x - 4 )
3 = A(x - 4 ) + B(x - 2 ) .......................................(1)
The aim now is to find the value of A and B. Note that if x = 4 , the term with A will be zero and if x = 2 the term with B will be zero.
let x = 4 in (1) : 3 = 2B# rArr B = 3/2 #
let x = 2 in (1) : 3 = -2A# rArr A = -3/2#
#rArr3/((x-2)(x-4)) = (3/2)/(x-4) - (3/2)/(x-2)# and the integral becomes :
#3/2intdx/(x-4) - 3/2intdx/(x-2) #
# = 3/2ln|x-4| - 3/2|x-2| + c#
