Question

How do you integrate `f(x)=1/((x-2)(x-5)(x+3))` using partial fractions?

Answer 1

`1/24ln|x-5| + 1/40ln|x+3| - 1/15ln|x-2| + c`

Explanation:

Since the factors on the denominator are linear then the numerators of the partial fractions will be constants , say A , B and C
`rArr 1/((x-2)(x-5)(x+3)) = A/(x-2) + B/(x-5) + C/(x+3)`

now, multiply through by (x-2)(x-5)(x+3)

so 1 = A(x-5)(x+3) + B(x-2)(x+3) + C(x-2)(x-5) ................................(1)

We now have to find the values of A , B and C. Note that if x = 2 , the terms with B and C will be zero. If x = 5 , the terms with A and C will be zero and if x = -3, the terms with A and B will be zero.
Making use of this fact , we obtain.

let x = 2 in (1) : 1 = -15A `rArr A = -1/15`

let x = 5 in (1) : 1 = 24B `rArr B = 1/24`

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

`rArr 1/((x-2)(x-5)(x+3)) = (-1/15)/(x-2) + (1/24)/(x-5) + (1/40)/(x+3)`

Integral now becomes.

`-1/15intdx/(x-2) + 1/24intdx/(x-5) + 1/40intdx/(x+3)`

`= 1/24ln|x-5| + 1/40ln|x+3| - 1/15ln|x-2| + c`