How to Integrate Partial Fraction Problems without a Prefix

In summary, the student is having trouble integrating the second term of an equation. They are looking for help from an expert on the topic and are receiving feedback.
  • #1
Youngster
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Homework Statement



[itex]\int[/itex][itex]\frac{8x^{2}+5x+8}{x^{3}-1}[/itex]

Homework Equations



Because the denominator can be reduced to (x-1)([itex]x^{2}+x+1[/itex]), I set up the partial fractions to be [itex]\frac{A}{(x-1)}[/itex] + [itex]\frac{Bx+C}{(x^{2}+x+1)}[/itex]

The Attempt at a Solution



I've solved for A, B, and C, and now have the integral set up as such:

7[itex]\int\frac{dx}{x-1}[/itex] + [itex]\int\frac{x-1}{x^{2}+x+1}[/itex]dx

Where A is 7, B is 1, and C is -1

I can integrate the first term simply, but I'm having trouble figuring out how to integrate the second term. The best I can think of is a u substitution, but du turns into 2x+1 dx, which is nothing like x-1 dx. Any suggestions?
 
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  • #2
Youngster said:

Homework Statement



[itex]\int[/itex][itex]\frac{8x^{2}+5x+8}{x^{3}-1}[/itex]

Homework Equations



Because the denominator can be reduced to (x-1)([itex]x^{2}+x+1[/itex]), I set up the partial fractions to be [itex]\frac{A}{(x-1)}[/itex] + [itex]\frac{Bx+C}{(x^{2}+x+1)}[/itex]

The Attempt at a Solution



I've solved for A, B, and C, and now have the integral set up as such:

7[itex]\int\frac{dx}{x-1}[/itex] + [itex]\int\frac{x-1}{x^{2}+x+1}[/itex]dx

Where A is 7, B is 1, and C is -1

I can integrate the first term simply, but I'm having trouble figuring out how to integrate the second term. The best I can think of is a u substitution, but du turns into 2x+1 dx, which is nothing like x-1 dx. Any suggestions?

Write x-1=(1/2)*(2x+1)-3/2. Now you can easily do the 2x+1 part. The -3/2 part is harder. You'll need to complete the square in the denominator and do a trig substitution.
 
  • #3
Complete the square on the other integral. I believe you'll get an arctan in the solution.
 
  • #4
Ah, I see now. It's been a while since I've done that, but it works. I suppose I should do similar exercises to get this in my head.

And yeah, part of the integral turned out to be an inverse tangent one. Thanks a lot.
 

FAQ: How to Integrate Partial Fraction Problems without a Prefix

1. What is a partial fraction problem?

A partial fraction problem is a mathematical problem that involves breaking down a complex rational expression into simpler fractions. This is done by finding the individual components, or partial fractions, that make up the original expression.

2. How do you solve a partial fraction problem?

To solve a partial fraction problem, you first need to factor the denominator of the rational expression. Then, you set up a system of equations using the coefficients of the partial fractions and solve for the unknown variables. Once you have the values for the variables, you can plug them back into the original expression to get the complete solution.

3. What are the different types of partial fraction problems?

There are two main types of partial fraction problems: proper and improper. Proper partial fraction problems have a numerator with a degree less than the denominator, while improper problems have a numerator with a degree equal to or greater than the denominator. Each type requires a different method of solving, but the overall process is the same.

4. Why are partial fraction problems important?

Partial fraction problems are important because they allow us to simplify complex rational expressions and make them easier to work with. This is particularly useful in calculus and other advanced math courses where these types of expressions are frequently encountered. Additionally, solving partial fraction problems can help develop problem-solving skills and improve understanding of algebraic concepts.

5. Can you use partial fractions to solve any type of rational expression?

No, partial fractions can only be used to solve rational expressions that can be factored. If the denominator cannot be factored, then the partial fraction method cannot be applied. Additionally, some rational expressions may require other techniques, such as completing the square or using the quadratic formula, to be solved.

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