How do you find int (11-2x)/(x^2 + x - 2) dx∫11−2xx2+x−2dx using partial fractions?

2 Answers
Narad T.
Oct 22, 2016

int((11-2x)dx)/((x-1)(x+2))=3ln(x-1)-5ln(x+2)+C∫(11−2x)dx(x−1)(x+2)=3ln(x−1)−5ln(x+2)+C

Explanation:

First factorise the denominator
x^2+x-2=(x-1)(x+2)x2+x−2=(x−1)(x+2)

Then we look for the partial fraction

(11-2x)/((x-1)(x+2))=A/(x-1)+B/(x+2)11−2x(x−1)(x+2)=Ax−1+Bx+2

=(A(x+2) +B(x-1))/((x-1)(x+2))=A(x+2)+B(x−1)(x−1)(x+2)

So 11-2x=A(x+2) +B(x-1)11−2x=A(x+2)+B(x−1)

If x=1x=1 =>⇒ 11-2=3A+011−2=3A+0
so A=9/3=3A=93=3

If x=-2x=−2 =>⇒ 11+4=0-3B11+4=0−3B
so B=15/-3=-5B=15−3=−5

(11-2x)/((x-1)(x+2))=3/(x-1)-5/(x+2)11−2x(x−1)(x+2)=3x−1−5x+2

Then we can integrate

int((11-2x)dx)/((x-1)(x+2))=int(3dx)/(x-1)-int(5dx)/(x+2)∫(11−2x)dx(x−1)(x+2)=∫3dxx−1−∫5dxx+2

=3ln(x-1)-5ln(x+2)+C=3ln(x−1)−5ln(x+2)+C

Frederico Guizini S.
Oct 23, 2016

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