Question

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

Answer 1

`[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`