Question

How do you integrate `int (2x+1)/((x+4)(x-1)(x+7))` using partial fractions?

Answer 1

`7/15ln|x+4| + 3/40ln|x-1| - 13/24ln|x+7| + c`

Explanation:

Since the factors on the denominator of the rational function are linear, then the numerators of the partial fractions will be constants, say A , B and C.

hence `(2x+1)/((x+4)(x-1)(x+7)) = A/(x+4) + B/(x-1) + C/(x+7)`

multiply through by (x+4)(x-1)(x+7) to obtain:

2x+1 = A(x-1)(x+7) + B(x+4)(x+7) + C(x+4)(x-1) ................(1)

The aim now is to find the value of A , B and C. Note that if x = -4 then the terms with B and C will be zero. If x = 1 the terms with A and C will be zero and if x =-7 the terms with A and B will be zero.
This is the starting point in finding A , B and C.

let x = -4 in (1) : -7 = -15A `rArr A = 7/15`

let x = 1 in (1) : 3 = 40B `rArr B = 3/40`

let x =-7 in (1) : -13 = 24C `rArr C = -13/24`

`(2x+1)/((x+4)(x-1)(x+7)) = (7/15)/(x+4) + (3/40)/(x-1) - (13/24)/(x+7)`

integral becomes:

`int(7/15)/(x+4) dx + int(3/40)/(x-1) dx - int(13/24)/(x+7) dx`

`= 7/15ln|x+4| + 3/40ln|x-1| - 13/24ln|x+7| + c`