How do you integrate int (x^3-x^2-5x )/( x^2-3x+2) using partial fractions?

1 Answer
Narad T.
Oct 29, 2016

The integral is =x^2/2+2x+5ln(x-1)-6ln(x-2)+C

Explanation:

As the degree of the numerator is greater than the denominator, let's do a long division
x^3-x^2-5xcolor(white)(aaaaaa)∣x^2-3x+2
x^3-3x^2+2xcolor(white)(aaaaa)∣x+2
0+2x^2-7x
color(white)(aaaa)2x^2-6x+4
color(white)(aaaaaa)0-x-4

We can factorise the denominator
x^2-3x+2=(x-1)(x-2)

So putting all together
(x^3-x^2-5x)/(x^2-3x+2)=x+2+(-x-4)/(x^2-3x+2)

Going to the decomposition into partial fractions
(-x-4)/(x^2-3x+2)=(-x-4)/((x-1)(x-2))=A/(x-1)+B/(x-2)
So -x-4=A(x-2)+B(x-1)
let x=1=>-5=-A so A=5
let x=2=>-6=B so B=-6
And we have
(x^3-x^2-5x)/(x^2-3x+2)=x+2+5/(x-1)-6/(x-2)
and the integral is
int((x^3-x^2-5x)dx)/(x^2-3x+2)=int(x+2+5/(x-1)-6/(x-2))dx
=x^2/2+2x+5ln(x-1)-6ln(x-2)+C

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