Question

How do you integrate `int(x+1)/((x-3)(x-1)(x+4))` using partial fractions?

Answer 1

`2/7ln|x-3| - 1/5ln|x-1| - 3/35ln|x+4| + c`

Explanation:

since the factors on the denominator are linear , the numerators will be constants.

`rArr (x+1)/((x-3)(x-1)(x+4)) = A/(x-3) + B/(x-1) + C/(x+4)`

multiply through by (x-3)(x-1)(x+4)

x+ 1 = A(x-1)(x+4) + B(x-3)(x+4) + C(x-3)(x-1).................(1)

now require to find values of A , B and C. Note that if x=1 , the terms with A and C will be zero. If x =3 , the terms with B and C will be zero and if x = -4 , the terms with A and B will be zero.
This is the starting point in finding values for A , B and C.

let x = 1 in (1) : 2 = - 10B → B =`-1/5`

let x = 3 in (1) : 4 = 14A → A `= 2/7`

let x = -4 in (1) : -3 = 35C → C`= -3/35`

`int((x+1))/((x-3)(x-1)(x+4)) dx =int( (2/7)/(x-3) - (1/5)/(x-1) - (3/35)/(x+4))dx`

`= 2/7ln|x-3| - 1/5ln|x-1| -3/35ln|x+4| + c`

where c, is the constant of integration.