
Help!
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! :)

This is a booger of an integral. You could expand into partial fractions and then integrate.
$\displaystyle \int\frac{1}{x^{3}+1}dx$$\displaystyle =\int\frac{1}{3(x+1)}dx$$\displaystyle +\int\frac{2}{3(x^{2}x+1)}dx\int\frac{x}{3(x^{2}x+1)}dx$
Two of can be tricky integrals.
$\displaystyle \int\frac{2}{3x^{2}3x+3}dx$
Let $\displaystyle x=u+\frac{1}{2};\;\ dx=du$
$\displaystyle 2\int\frac{1}{3(u+\frac{1}{2})^{2}3(u+\frac{1}{2})+3}dx$
$\displaystyle 2\int\frac{1}{3u^{2}+\frac{9}{4}}dx$
Multiply by 4:
$\displaystyle 8\int\frac{1}{12u^{2}+9}dx$
You may want to try another substitution, say, $\displaystyle tan^{1}(\frac{2}{3}u\sqrt{3})$
Anyway, this particular one works out to:
$\displaystyle \frac{4\sqrt{3}tan^{1}(\frac{\sqrt{3}(2x1)}{3})}9}$
There's a head start. Of course, using your previous solutions will help. I done it from scratch.

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 nonrepeated 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.

OMG this question is killing me. I still can't get the partial fractions part. Do you think you could do a stepbystep for me for that?

Hello Margarita.
$\displaystyle x^{3}+1$ factors into $\displaystyle (x+1)(x^{2}x+1)$
This gives the PFD of $\displaystyle \frac{A}{x+1}+\frac{Bx+C}{x^{2}x+1}=1$
$\displaystyle A(x^{2}x+1)+(Bx+C)(x+1)=1$
$\displaystyle Ax^{2}Ax+A+Bx^{2}+Bx+Cx+C=1$
$\displaystyle A+B=0;\;\ A+B+C=0;\;\ A+C=1$
This leads to $\displaystyle A=\frac{1}{3};\;\ B=\frac{1}{3};\;\ C=\frac{2}{3}$

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!

I edited my post. Just think, after you get the partial fraction, you have the
insidious integration to contend with. :D