How do you integrate #int(x)/((3x-2)(x+2)(2x-1))# using partial fractions?

1 Answer
Feb 18, 2016

#1/4ln|3x-2| - 1/20ln|x+2| -1/5ln|2x-1| + c#

Explanation:

Since the factors on the denominator are linear , the numerators will be constants , say A , B and C.

#x/((3x-2)(x+2)(2x-1)) = A/(3x-2) + B/(x+2) + C/(2x-1)#

Now multiply through by (3x-2)(x+2)(2x-1)

hence : x = A(x+2)(2x-1) + B(3x-2)(2x-1) + C(3x-2)(x+2)...(1)

The aim now is to find values of A , B and C. Note that if x =-2 , the terms with A and C will be zero.
If x#=1/2" the terms with A and B will be zero"#
and if x#=2/3"the terms with B and C will be zero"#

let x = -2 in (1): - 2 = 40B #rArr B =-1/20#
let #x = 1/2" in (1)" : 1/2 = -5/4C rArr C= -2/5 #
let # x = 2/3 " in (1)": 2/3 = 8/9A rArr A = 3/4 #

hence integral becomes

#int(3/4)/(3x-2) dx -(1/20)/(x+2) dx -(2/5)/(2x-1) dx #

# = 3/4 . 1/3 ln|3x-2| -1/20ln|x+2| -2/5 . 1/2 ln|2x-1| + c#

# = 1/4ln|3x-2| - 1/20ln|x+2| - 1/5ln|2x-1| + c#