Question

How do you integrate `int (x^2 +x -16) / [(x+1)(x-3)^2] dx` using partial fractions?

Answer 1

`int \ (x^2+x-16)/((x+1)(x-3)^2) \ dx = -ln|x+1| + 2 ln|x-3| + 1/(x-3) + C`

Explanation:

We have:

`I = int \ (x^2+x-16)/((x+1)(x-3)^2) \ dx`

We can decompose the integrand into partial fraction, which will take the form:

`(x^2+x-16)/((x+1)(x-3)^2) -= A/(x+1)+B/(x-3) + C/(x-3)^2`
`" " = (A(x-3)^2+B(x+1)(x-3) + C(x+1))/((x+1)(x-3)^2)`

Leading to the identity:

`x^2+x-16 -= A(x-3)^2+B(x+1)(x-3) + C(x+1)`

Where `A,B,C` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x=-1 => -16=A(-4)^2 => A=-1`
Put `x= \ \ \ \ 3 => -4=4C=> C=-1`
Put `x= \ \ \ \ 0 => -16=9A-3B+C => B=2`

So using partial fraction decomposition we have:

`I = int \ -1/(x+1)+2/(x-3) - 1/(x-3)^2 \ dx`
`\ \ = int \ -1/(x+1) \ dx +2 \ int \ 1/(x-3) \ dx - int \ 1/(x-3)^2 \ dx`

And now all integrands are readily integratable, so:

`I = -ln|x+1| + 2 ln|x-3| + 1/(x-3) + C`