1. ## 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!

2. This is a booger of an integral. You could expand into partial fractions and then integrate.

$\int\frac{1}{x^{3}+1}dx$ $=\int\frac{1}{3(x+1)}dx$ $+\int\frac{2}{3(x^{2}-x+1)}dx-\int\frac{x}{3(x^{2}-x+1)}dx$

Two of can be tricky integrals.

$\int\frac{2}{3x^{2}-3x+3}dx$

Let $x=u+\frac{1}{2};\;\ dx=du$

$2\int\frac{1}{3(u+\frac{1}{2})^{2}-3(u+\frac{1}{2})+3}dx$

$2\int\frac{1}{3u^{2}+\frac{9}{4}}dx$

Multiply by 4:

$8\int\frac{1}{12u^{2}+9}dx$

You may want to try another substitution, say, $tan^{-1}(\frac{2}{3}u\sqrt{3})$

Anyway, this particular one works out to:

$\frac{4\sqrt{3}tan^{-1}(\frac{\sqrt{3}(2x-1)}{3})}9}$

There's a head start. Of course, using your previous solutions will help. I done it from scratch.

3. 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.

4. 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?

5. Hello Margarita.

$x^{3}+1$ factors into $(x+1)(x^{2}-x+1)$

This gives the PFD of $\frac{A}{x+1}+\frac{Bx+C}{x^{2}-x+1}=1$

$A(x^{2}-x+1)+(Bx+C)(x+1)=1$

$Ax^{2}-Ax+A+Bx^{2}+Bx+Cx+C=1$

$A+B=0;\;\ -A+B+C=0;\;\ A+C=1$

This leads to $A=\frac{1}{3};\;\ B=\frac{-1}{3};\;\ C=\frac{2}{3}$

6. 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!

7. I edited my post. Just think, after you get the partial fraction, you have the

insidious integration to contend with.