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

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