# Find the indefinite integral using integration by partial fractions: S 6/(x^3 -3x^2) dxI find partial fractions difficult, but have possibly figured the following out: 6/[x^2(x-3)] dx...

Find the indefinite integral using integration by partial fractions: S 6/(x^3 -3x^2) dx

I find partial fractions difficult, but have possibly figured the following out:

6/[x^2(x-3)] dx

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

6=Ax^2(x-3)+Bx(x-3)+C(x-3)

Letting x=0

C=-4

but I'm not sure that is right, I have been unable to find any explenation on how to do a problem with 2 exponents in the denominator.

As always, any help would be greatly appreciated.

### 1 Answer | Add Yours

You need to use partial fraction decomposition such that:

`6/(x^2(x-3)) = A/x + B/(x^2) + C/(x-3)`

Bringing the terms to the right to a common denominator yields:

`6 = A(x(x-3)) + B(x-3) + Cx^2`

`6 = Ax^2 - 3Ax + Bx - 3B + Cx^2`

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

Equating the coefficients of like powers yields:

`A+C = 0 => A=-C`

`-3A+B = 0 => -3A = -B => 3A = B`

`-3B = 6 => B = -2 > A = B/3 => A = -2/3 => C = 2/3`

`6/(x^2(x-3)) = -2/(3x)- 2/(x^2) + 2/(3(x-3))`

Integrating both sides yields:

`int 6/(x^2(x-3)) dx= int -2/(3x) dx- int 2/(x^2) dx+ int 2/(3(x-3)) dx`

`int 6/(x^2(x-3)) dx = -(2/3)ln|x| - 2 int x^(-2)dx + (2/3)ln|x-3| `

`int 6/(x^2(x-3)) dx = -(2/3)ln|x| + 2/x + (2/3)ln|x-3| + c`

`int 6/(x^2(x-3)) dx = (2/3)ln|(x-3)/x| + 2/x + c`

**Hence, evaluating the indefinite integral using partial fraction decomposition,Â yields `int 6/(x^2(x-3)) dx = (2/3)ln|(x-3)/x| + 2/x + c.` **