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

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lfryerda | High School Teacher | (Level 2) Educator

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




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



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:


`1/4=b+c`  multiply by 4

`1=4b+4c`   equation (2)

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



which means that 


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


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

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