evaluate the integral integrate of ((3)/(x^3-x^2-12x))dx
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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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