How do you properly solve the integral ∫(6/(x^2 - 4x -12).dx?

[brandy]

Homework Statement

∫(6/(x^2 - 4x -12).dx

The Attempt at a Solution

i went
6*∫1/(x^2 - 4x -12)
which i then worked out to be 6*ln(x^2 - 4x -12)/(2x-4) +c
ie 6*ln u / u'
where u=x^2 - 4x -12

im not very good at this. is this correct?

 

[HallsofIvy]

No that is not the integral. I imagine you tried making the substitution u= x^2- 4x- 12 so that du= (2x-4). But you cannot just divide by 2x-4 and say that dx= du/(2x-4) because that is a function of x itself- you cannot just move it inside the integral sign as you could a constant. Instead factor x^2- 4x- 12 and use "partial fractions".

 

[brandy]

i don't really know what you are saying. can you dumb it down a lot more. I am a real dummy :)

my thinking:
6/(x^2 - 4x -12). 6 is a constant so it can be brought to the other side of the integral.
now you have 6*∫1/(x^2 - 4x -12)
i thought that ∫1/(x^2 - 4x -12) is ln (x^2 - 4x -12)/(2x-4)
i take it this is where I am wrong?
can you explain better?

 

[Cyosis]

Using your substitution you would get the following integral:

$$6 \int \frac{1}{2x-4} \frac{du}{u}$$

You then seem to treat the $$\frac{1}{2x-4}$$ term as a constant and you put it in front of the integral and integrate with respect to u. The mistake you make here is that x itself depends on u, so x is not a constant with respect to u. So before you integrate you have to rewrite the fraction in terms of u. As a result you can't take the fraction out of the integral.

As Ivy suggested you want to factor the denominator and then write the integrand as $$\frac{A}{x+a}+\frac{B}{x+b}$$. You should be able to determine the constants A,B,a and b and then solve the integral.

On a different note if you're unsure whether or not you did the integration correctly, it is a good habit to take the derivative of your answer and see if it equals the integrand. If it doesn't you know you have made a mistake.