Solve Integral: ∫ (5x^2 - 42x + 24)/(x^3 - 10x^2 + 12x + 72) dx

[Pyroadept]

Homework Statement

Q. Solve ∫ (5x^2 - 42x + 24)/(x^3 - 10x^2 + 12x + 72) dx

Homework Equations

∫ x^n dx = ∫ (1/n+1)x^(n+1) + C
∫ 1/x dx = ln x + C

The Attempt at a Solution

Hi everyone,

Here's what I've done so far:

(i) I factored the bottom equation into (x + 2)(x - 6)^2

and then separated the fraction 1/[(x + 2)(x - 6)^2] into a sum of three fractions of the form:
A/(x + 2) + B/(x - 6) + C/[ (x-6)^2 ]

and found that A = 1/64, B = -1/64, C = 1/8

(ii) The only way I can see to go from here is to split the top integral into three integrals:

(1/64) ∫ (5x^2 - 42x + 24)/(x + 2) dx + (-1/64) ∫ (5x^2 - 42x + 24)/(x - 6) dx + (1/8) ∫ (5x^2 - 42x + 24)/(x-6)^2 dx

and solve them all individually using substitution, letting u = x+2, x-6 and x-6 respectively.

The answer I get in this way is: (5/8)x - 21/4 +2ln(x+2) + 3ln(x-6) + 6/(x-6) + C, C = constant of integration,

which is incorrect, according to the WebWork site, but I can't find any algebraic errors.
So, I'm thinking perhaps what I do in step (ii) is not the best way to do it.

Has anyone any suggestions?

Thanks for any help!

 

[Dick]

You don't want to do it that way. You want to split (5x^2 - 42x + 24)/((x+2)*(x-6)^2) into three fractions of the form A/(x + 2) + B/(x - 6) + C/[ (x-6)^2 ] with A, B and C constant.

 

[Cyosis]

You have split your fractions correctly. However splitting $1/((x-6)^2(x+2))$ instead of the entire integrand, like Dick suggests, results in you having to integrate three integrals which are very sensitive to mistakes. I checked the expression you wrote down in (ii) in mathematica and it does give you the correct answer so you must have made a mistake integrating somewhere. However I suggest you do it Dick's way, it gives easy values for A, B and C (integers) and results in three very easy integrals.