To solve the integral, we need to convert the integrand into a set of partial fractions. That is, consider:

`3/{x^3-x^2-12x}`

`=3/{x(x^2-x-12)}`

`=3/{x(x-4)(x+3)}`

`=a/x+b/{x-4}+c/{x+3}` Now expand and compare with coefficients

`3=a(x-4)(x+3)+bx(x+3)+cx(x-4)`

`3=a(x^2-x-12)+b(x^2+3x)+c(x^2-4x)`

This means that comparing the constant coefficients gives:

`3=-12a` so `a=-1/4`

Comparing linear terms:

`0=-a+3b-4c` to give the equation:

`1/4=3b-4c` and multiply by 4

`1=12b-16c` equation (1)

The quadratic terms:

`0=a+b+c`

`1/4=b+c` multiply by 4

`1=4b+4c` equation (2)

Now consider 4x(2)+(1) to get:

`5=28b`

`b=5/28`

which means that

`c=1/14`

Therefore, the integral becomes:

`int 3/{x^3-x^2-12x}dx`

`=-1/4int 1/xdx+5/28int1/{x-4}dx+1/14int1/{x+3}dx`

`=-1/4lnx+5/28ln(x-4)+1/14ln(x+3)+C` where C is the constant of integration

`=ln({(x-4)^{5/28}(x+3)^{1/14}}/x^{1/4})+C`

**The integral evaluates to `ln({(x-4)^{5/28}(x+3)^{1/14}}/x^{1/4})+C` .**

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