Yes I'm still at integration lol but this is the last question for the book so anyway the question's:
The indefinite integral (integrate) [P(x)/(x^3 + 1)] dx, where P(x) is a polynomial in x, is denoted by I.
(i) Find I when P(x) = x^2. (Answer: 1/3 ln | x^3 + 1 | + C)
(ii) By writing x^3 + 1 = (x + 1)(x^2 + Ax + B), where A and B are constants, find I when
(a) P(x) = x^2 - x + 1, (Answer: ln | x + 1 | + C)
(b) P(x) = x + 1 (Answer: 2/(sq root 3) tan^-1 2/(sq root 3) (x - 1/2) + C)
(iii) Using the results of parts (i) and (ii), or otherwise, find I when P(x) = 1. (Answer: 1/3 ln | x + 1 | + 1/6 ln | x^2 - x + 1 | + 1/(sq root 3) tan^-1 2/(sq root 3) (x - 1/2) + C)
I got the answers for all except the last part (iii).
So thanks if anyone could help! :)
Jul 31st 2006, 06:22 AM
This is a booger of an integral. You could expand into partial fractions and then integrate.
Two of can be tricky integrals.
Multiply by 4:
You may want to try another substitution, say,
Anyway, this particular one works out to:
There's a head start. Of course, using your previous solutions will help. I done it from scratch.
Jul 31st 2006, 06:40 AM
I can't quite get the first part on partial fractions. They don't look familiar.
EDIT: I just looked through my formula list, is the partial fraction done by non-repeated quadratic factor? I think so but I'm not sure, I'll give it a try. Thanks for your help by the way, I forgot to thank you just now.
Jul 31st 2006, 06:53 AM
OMG this question is killing me. I still can't get the partial fractions part. Do you think you could do a step-by-step for me for that?
Jul 31st 2006, 06:58 AM
This gives the PFD of
This leads to
Jul 31st 2006, 07:01 AM
I'm there now but I'm having problems finding B and C. I only got A which is 1/3?
EDIT: Thanks so much, galactus, I'm such a noob lol I actually went to separate the B and C. That's why I can't find the answers. I should have no problems now I think so thanks once again!
Jul 31st 2006, 07:07 AM
I edited my post. Just think, after you get the partial fraction, you have the