How do you find int (3x^2-x)/((1 - x)^2(1 - 3x))dx using partial fractions?

1 Answer
Nov 26, 2015

[ln(1/(x-1))+1/(x-1)]+C

Explanation:

int (3x^2-x)/((1 - x)^2(1 - 3x))dx

EEA,EEB,EEC such as

-(3x^2-x)/((x-1)^2(3x-1)) = -A/(x-1)-B/(x-1)^2-C/(3x-1)

(I factorized the denominator to do partial fraction)

Multiply both side by : (x-1)^2(3x-1)

3x^2-x = A(3x-1)(x-1)+ B(3x-1) + C(x-1)^2

3x^2-x = A(1-4x+3x^2) + B(3x-1) + C(1+x^2-2x)

3x^2-x = A-4Ax+3Ax^2-B+3Bx+C+Cx^2-2Cx

3x^2-x = x^2(C+3A)+x(-4A+3B-2C) + A - B + C

by identification

C+3A = 3
-4A+3B-2C = -1
A-B+C = 0

you find ( ;) )

A = 1
B = 1
C = 0

and then

int(3x^2-x)/((x-1)^2(3x-1))dx = -int1/(x-1) + 1/(x-1)^2dx

Which is [ln(1/(x-1))+1/(x-1)]+C

(Because aln(b) = ln(b^a))

and -int1/u^2 = 1/u