How do you express (x-1)/(x^3 +x^2)x1x3+x2 in partial fractions?

1 Answer
Oct 29, 2016

The answer is (x-1)/(x^3+x^2)=2/x-1/x^2-2/(x+1)x1x3+x2=2x1x22x+1

Explanation:

Some factorisation to start with
(x-1)/(x^3+x^2)=(x-1)/((x^2)(x+1))x1x3+x2=x1(x2)(x+1)
=A/x+B/x^2+C/(x+1)=Ax+Bx2+Cx+1
=(Ax(x+1)+B(x+1)+Cx^2)/((x^2)(x+1))=Ax(x+1)+B(x+1)+Cx2(x2)(x+1)

So now we can solve for A, B,and CA,B,andC
x-1=Ax(x+1)+B(x+1)+Cx^2x1=Ax(x+1)+B(x+1)+Cx2
let x=-1, -2=C2=C=>C=-2C=2
Coefficients of x^2x2, 0=A+C0=A+C=>A=2A=2
Coefficients of xx, 1=A+B1=A+B=>B=-1B=1

And finally, we have
(x-1)/(x^3+x^2)=2/x-1/x^2-2/(x+1)x1x3+x2=2x1x22x+1