Calc II - Integration of Partial Fractions

In summary, the partial fraction problem is confusing because the numerator in one term is wrong. You need to use coefficients to solve the problem.
  • #1
demersal
41
0

Homework Statement


Hi everyone, here is a new partial fractions question I just cannot understand:

[tex]\int[/tex][tex]\frac{x^{3}}{x^{3}+1}[/tex]dx


Homework Equations



Partial Fractions, difference of perfect cubes, polynomial long division

The Attempt at a Solution



[tex]\int[/tex][tex]\frac{x^{3}}{x^{3}+1}[/tex] dx

[tex]\int[/tex]1 dx + [tex]\int[/tex][tex]\frac{-1}{x^{3}+1}[/tex] dx

x + [tex]\int[/tex][tex]\frac{-1}{(x+1)(x^{2}-x+1)}[/tex] dx

[tex]\frac{A}{x+1}[/tex] + [tex]\frac{B}{(x^{2}-x+1)}[/tex]

A(x[tex]^{2}[/tex]-x+1) + B(x+1) = - 1

If I use coefficients:
x[tex]^{2}[/tex]: 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!
 
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  • #2
[tex]\frac{A}{x+1}+\frac{B}{(x^{2}-x+1)}[/tex]

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
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
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
The numerator of each term must be a polynomial of degree one less than the degree of the denominator. [tex]x^2-x+1[/tex] is a 2nd degree polynomial, so the numerator for that term should be a first degree polynomial; [tex]Bx+C[/tex].
 
  • #6
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
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
So it should be Bx?
 
  • #9
ohh I see gabbagabbahey, I should have a Bx and a separate C over the same denominator?
 
  • #10
demersal said:
ohh I see gabbagabbahey, I should have a Bx and a separate C over the same denominator?

You should have:
[tex]\frac{A}{x+1}+\frac{Bx+C}{(x^{2}-x+1)} [/tex]
 
  • #11
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
[tex]Ax^2+Bx+C[/tex].
You must include all the lower order terms too (i.e. the 2nd (A), 1st (B) and 0th (C) terms).
 
  • #12
Ok! I got it now! Thank you both! Hopefully I'll never have to post about partial fractions again :)
 

1. What is partial fraction decomposition?

Partial fraction decomposition is a method of breaking down a rational function into smaller, simpler fractions. This method is used in Calculus II for integration purposes.

2. Why do we use partial fraction decomposition?

Partial fraction decomposition allows us to simplify complex rational functions, making it easier to integrate them. It also helps us to find the antiderivative of a rational function, which is necessary for finding definite integrals.

3. How do you perform partial fraction decomposition?

To perform partial fraction decomposition, we first factor the denominator of the rational function. Then, we write the function as a sum of smaller fractions with each factor of the denominator as the denominator of one of the fractions. We then solve for the unknown coefficients using algebraic methods.

4. What is the difference between proper and improper fractions in partial fraction decomposition?

A proper fraction is a fraction where the degree of the numerator is less than the degree of the denominator. In partial fraction decomposition, proper fractions are relatively easy to decompose. An improper fraction, on the other hand, has a degree of the numerator greater than or equal to the degree of the denominator. These fractions require additional steps to decompose.

5. Are there any limitations to using partial fraction decomposition?

Partial fraction decomposition can only be used for rational functions, which means that the numerator and denominator must both be polynomials. It also cannot be used for functions with repeated factors in the denominator. In these cases, alternative methods of integration may need to be used.

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