# Calc II - Integration of Partial Fractions

1. Sep 21, 2008

### demersal

1. The problem statement, all variables and given/known data
Hi everyone, here is a new partial fractions question I just cannot understand:

$$\int$$$$\frac{x^{3}}{x^{3}+1}$$dx

2. Relevant equations

Partial Fractions, difference of perfect cubes, polynomial long division

3. The attempt at a solution

$$\int$$$$\frac{x^{3}}{x^{3}+1}$$ dx

$$\int$$1 dx + $$\int$$$$\frac{-1}{x^{3}+1}$$ dx

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

$$\frac{A}{x+1}$$ + $$\frac{B}{(x^{2}-x+1)}$$

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

If I use coefficients:
x$$^{2}$$: 0=A
x: 0 = B-A
: -1 = A+B

These don't add up! Also, using critical values leads me to a similar problem. What am I doing wrong in my strategy of attacking these partial fractions??

Thank you for your time and help, you all are so wonderful!

2. Sep 21, 2008

### Pacopag

$$\frac{A}{x+1}+\frac{B}{(x^{2}-x+1)}$$

Remember that when you decompose the fraction, the numerator must be a general polynomial with degree one less than that of the denominator. What I'm getting at is that the numerator in your "B"-term is wrong. What should it be instead?

3. Sep 21, 2008

### demersal

Hmm, I don't quite understand what you are saying. I thought I was just supposed to multiply the numerator by the denominator and cancel. Is there another concept I am missing?

4. Sep 21, 2008

### statdad

If the denominator is linear, the numerator must be a constant. If the denominator is an irreducible quadratic, the numerator should be a ********** (hint: not a constant)

5. Sep 21, 2008

### gabbagabbahey

The numerator of each term must be a polynomial of degree one less than the degree of the denominator. $$x^2-x+1$$ is a 2nd degree polynomial, so the numerator for that term should be a first degree polynomial; $$Bx+C$$.

6. Sep 21, 2008

### Pacopag

In your second term, you just put a B in the numerator. But that is a zeroth order polynomial when you have a second order polynomial in the denominator. So instead of just B, you need to put in an arbitrary first order polynomial. What is another name for a first order polynomial?

7. Sep 21, 2008

### Pacopag

Right gabbagabbahey. Notice that the largest exponent in the denominator is 2. So the largest exponent in the numerator should be 1 (i.e. one less than 2).

8. Sep 21, 2008

### demersal

So it should be Bx?

9. Sep 21, 2008

### demersal

ohh I see gabbagabbahey, I should have a Bx and a separate C over the same denominator?

10. Sep 21, 2008

### gabbagabbahey

You should have:
$$\frac{A}{x+1}+\frac{Bx+C}{(x^{2}-x+1)}$$

11. Sep 21, 2008

### Pacopag

Yes. By "general" or "arbitrary" we mean that you have to include all the terms. e.g. for a second order polynomial you'd have
$$Ax^2+Bx+C$$.
You must include all the lower order terms too (i.e. the 2nd (A), 1st (B) and 0th (C) terms).

12. Sep 21, 2008

### demersal

Ok! I got it now! Thank you both! Hopefully I'll never have to post about partial fractions again :)

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