Question

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

Answer 1

`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`