Partial Fractions: Integrating a Problem

In summary, the conversation discusses a problem involving integrating using partial fractions. The attempt at a solution involved using long division to get the integral of (x3-4x2+x+6)/(x2-x+2) and splitting it into three separate integrals. The question then asked for suggestions on how to proceed from there. The response provided guidance on how to approach the integrals and suggested completing the square in the denominator for one of them.
  • #1
schapman22
74
0

Homework Statement



We were given a worksheet to integrate some problems using partial fractions. This one however I cannot figure out what to do with it. This is the problem.

∫(x3-4x2+x+6)/(x2-x+2) dx

The Attempt at a Solution



using long division i got ∫(x-3) (-4(x-3)/(x2-x+2) dx
split it up into ∫xdx - 3∫dx - 4∫(x-3)/(x2-x+2)
now I don't know what to do with the x2-x+2?
any suggestions on how to proceed from here? Thanks in advance!
 
Physics news on Phys.org
  • #2
schapman22 said:

Homework Statement



We were given a worksheet to integrate some problems using partial fractions. This one however I cannot figure out what to do with it. This is the problem.

∫(x3-4x2+x+6)/(x2-x+2) dx

The Attempt at a Solution



using long division i got ∫(x-3) + (-4(x-3)/(x2-x+2) dx
split it up into ∫xdx - 3∫dx - 4∫(x-3)/(x2-x+2)
now I don't know what to do with the x2-x+2?
any suggestions on how to proceed from here? Thanks in advance!
What you have done so far looks correct.

For the integral [itex]\displaystyle \int 2\frac{2x-6}{x^2-x+2}\,dx\,,[/itex] notice that the derivative of the denominator is 2x-1.

Split the integrand up into [itex]\displaystyle
2\frac{2x-1}{x^2-x+2}+2\frac{-5}{x^2-x+2}\,.[/itex]

So you get two integrals. The first results in a logarithm. For the second one, complete the square in the denominator.
 
  • #3
So I'm allowed to multiple the integrand by 2?
 
  • #4
thank you
 

Related to Partial Fractions: Integrating a Problem

1. What are partial fractions?

Partial fractions are a method used in calculus to simplify the integration of a complex rational function into smaller, easier-to-solve fractions.

2. When is it necessary to use partial fractions?

Partial fractions are typically used when integrating a rational function that cannot be easily integrated using other methods, such as substitution or integration by parts.

3. How do you find the partial fraction decomposition of a rational function?

To find the partial fraction decomposition, the rational function is first factored into its irreducible factors. Then, the coefficients of each factor are solved for using a system of equations.

4. What is the purpose of using partial fractions?

The purpose of using partial fractions is to simplify the integration process of a complex rational function. By breaking it down into smaller fractions, it becomes easier to integrate and can lead to a more accurate solution.

5. Are there any limitations to using partial fractions?

Yes, partial fractions can only be used for rational functions, meaning they must have a polynomial in the numerator and denominator. Additionally, the degree of the numerator must be less than the degree of the denominator for the method to work.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
985
  • Calculus and Beyond Homework Help
Replies
4
Views
791
  • Calculus and Beyond Homework Help
Replies
3
Views
411
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
293
  • Calculus and Beyond Homework Help
Replies
25
Views
486
Replies
9
Views
770
  • Calculus and Beyond Homework Help
Replies
10
Views
539
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top