You need to use partial fraction decomposition to write the given integral as a sum (difference) of a simpler integrals such that:

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

`1 = Ax(x-2) + B(x-2) + Cx^2`

`1 = Ax^2 - 2Ax + Bx - 2B + Cx^2`

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

Equating the coefficients of like powers yields:

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

`-2A + B = 0 `

`-2B = 1 => B = -1/2`

Substituting `-1/2` for B in `-2A + B = 0 ` yields:

`-2A- 1/2 = 0 => -2A = 1/2 => A = -1/4 => C = 1/4`

`1/(x^2(x-2)) = -1/(4x) - 1/(2x^2) + 1/(4(x-2))`

Integrating both sides yields:

`int 1/(x^2(x-2)) dx= int -1/(4x) dx- int 1/(2x^2) dx+ int 1/(4(x-2))dx`

`int 1/(x^2(x-2)) dx =-(1/4)ln|x| -(1/2)(x^(-2+1))/(-2+1) + int 1/(4t)dt`

`int 1/(x^2(x-2)) dx = -(1/4)ln|x| + 1/(2x) + (1/4)ln|t| + c`

Substituting back `x - 2` for `t` yields:

`int 1/(x^2(x-2)) dx = -(1/4)ln|x| + 1/(2x) + (1/4)ln|x - 2| + c`

**Hence, evaluating the given integral, using partial fraction decomposition, yields `int 1/(x^2(x-2)) dx = -(1/4)ln|x| + 1/(2x) + (1/4)ln|x - 2| + c.` **

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