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

1 Answer
Oct 9, 2016

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