Question

How do you integrate `int (x-x^2)/((x+3)(x-6)(x+4))` using partial fractions?

Answer 1

`int f(x) = -302/5 ln|x+3| -39/5 ln|x-6| +356/5 ln|x+4| +C`

Explanation:

First split the expression up into three elements, one for each of the bracketed terms in the denominator.

`f(x) = (x-x^2)/((x+3)(x-6)(x+4))`
`f(x) = A/(x+3) + B/(x-6) + C/(x+4)`

Restoring the denominator gives us
`f(x) =( A(x-6)(x+4) + B(x+3)(x+4) + C(x+3)(x-6))/((x+3)(x-6)(x+4))`
`f(x)=(Ax^2 -2Ax -24A +Bx^2+7Bx +12B +Cx^2-3Cx-18C)/((x+3)(x-6)(x+4))`

`= ((A+B+C)x^2 +(-2A+7B-3C)x +(-24A +12B-18C))/((x+3)(x-6)(x+4))`

Each of the coefficients in the numerator must equal the coefficients in the original equation. Therefore
`A+B+C = -1`
`-2A+7B-3C = 1`
`-24A+12B-18C = 0`

We can now solve these three equations to find A, B, and C.
Doubling the first equation and adding it to the second gives
`9B-C =-1` or `C = 9B+1`

Multiplying the second equation by 12 and subtracting it from the third gives
`-72B+18C = -12` or `12B+3C = -2`

Substituting for C give `12B +3(9B+1) = -2`
`39B = -5`
`B = -39/5`
`C = 9B+1 =356/5`
`A = (-5+39-356)/5 = -302/5`

We can therefore write the original equation as
`F(x) = -302/(5(x+3)) - 39/(5(x-6)) + 356/(5(x+4))`

`int f(x) = -302/5 ln|x+3| -39/5 ln|x-6| +356/5 ln|x+4| +C`