How do you use partial fraction decomposition to decompose the fraction to integrate (x^2-x+2)/(x(x-1)^2)x2−x+2x(x−1)2?

1 Answer
Jim H
Jul 27, 2015

(x^2-x+2)/(x(x-1)^2) = 2/x-1/(x-1)+2/(x-1)^2x2−x+2x(x−1)2=2x−1x−1+2(x−1)2

Explanation:

(x^2-x+2)/(x(x-1)^2) = A/x+B/(x-1)+C/(x-1)^2x2−x+2x(x−1)2=Ax+Bx−1+C(x−1)2

Multiply both sides by x(x-1)^2x(x−1)2 (or get a common denominator and compare the numerators -- you'll get to the same place, but one approach may be clearer to you)

A(x-1)^2+Bx(x-1)+Cx = x^2-x+2A(x−1)2+Bx(x−1)+Cx=x2−x+2

Ax^2-2Ax+1+Bx^2-Bx+Cx = x^2-x+2Ax2−2Ax+1+Bx2−Bx+Cx=x2−x+2

[A+B]x^2 + [-2A-B+C]x +[A] = 1x^2-1x+2[A+B]x2+[−2A−B+C]x+[A]=1x2−1x+2

So we get:

A+B = 1A+B=1
-2A+B+C=-1−2A+B+C=−1
A = 2A=2

Eq 1 and 3 get us A=2A=2 and B = -1B=−1

Substituting in Eq 2, we get C = 2C=2

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