Question

How do you integrate `(x - 16) / (x^2 + x - 2)` using partial fractions?

Answer 1

`int((x-16)/(x^2+x-2))dx = 6 ln|x+2|-5ln|x-1|+c`

Explanation:

We first expand the given expression into partial fractions:

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

And so. `(x-16)-=A(x-1)+B(x+2)`

Put `x=1=>1-16=0+B(3)`
`:. 3B=-15=>B=-5`

Put `x=-2=>-2-16=A(-2-1)+0`
`:. -3A=-18=>A=6`

So the partial fraction decomposition is:
`(x-16)/(x^2+x-2)-=6/(x+2)-5/(x-1)`

We now want to integrate; so
`int((x-16)/(x^2+x-2))dx = int(6/(x+2)-5/(x-1))dx`
`int((x-16)/(x^2+x-2))dx = 6 int(1/(x+2))dx-5int(1/(x-1))dx`
`int((x-16)/(x^2+x-2))dx = 6 ln|x+2|-5ln|x-1|+c`