How do you integrate #(2x-1)/(x^2-x-6)# using partial fractions?

1 Answer
Oct 29, 2016

THe answer is #=ln(x+2)+ln(x-3)+C#

Explanation:

Let's factorise the denominator
#(2x-1)/(x^2-x-6)=(2x-1)/((x+2)(x-3))#
Then we can decompose into partial fractions
Let #(2x-1)/((x+2)(x-3))=A/(x+2)+B/(x-3)#
#=(A(x-3)+B(x+2))/((x+2)(x-3))#
so we can compare the numerators
#2x-1=A(x-3)+B(x+2)#
let #x=3##=>##5=5B##=>##B=1#
let #x=-2##=>##-5=-5A##=>##A=1#
and finally we have
#(2x-1)/((x+2)(x-3))=1/(x+2)+1/(x-3)#
Now we can integrate
#int((2x-1)dx)/((x+2)(x-3))=intdx/(x+2)+intdx/(x-3)#
#=ln(x+2)+ln(x-3)+C#