Help with Partial Fraction Decomposition

In summary: Keep going with the trig identities to simplify the integral further. In summary, to solve the integral \int 1/(x^{2}+4x+5)^{2} using partial fraction decomposition, the denominator must have distinct factors. If the denominator has only one type of polynomial squared or cubed, then partial fractions cannot be used. In this case, the denominator can be rewritten in the form (x+a)^2+b^2, and a trig substitution can be used. After completing the square and substituting u for x+a, the integral becomes du/(u^2+1)^2. Then, using the substitution u = tan(theta) and trig identities, the integral can be simplified to 1/sec^2(theta
  • #1
akbar786
18
0

Homework Statement



[tex]\int[/tex] 1/(x[tex]^{2}[/tex]+4x+5)[tex]^{2}[/tex]

Homework Equations



I am using partial fraction decomposition

The Attempt at a Solution



1/(x[tex]^{2}[/tex]+4x+5)[tex]^{2}[/tex] = Ax+b/(X^2+4x+5) + Cx+D/(x^2+4x+5)^2

1 = (Ax+b)(X^2+4x+5) + Cx+D

When i multiply through to find the values for A and b and c i get zero because there is no X^3 or x^2 or x on the left side. Am i doing something wrong or do i need to use another method to solve this integral?
 
Physics news on Phys.org
  • #2
The integrand is already in expanded form. You need to use another method. Start by rewriting the denominator in the form (x+a)2+b2, and then use a trig substitution.
 
  • #3
so when there is only dx sitting on the top i cannot use partial fraction decomposition? How can i tell if its already in expanded form. On the directions of this problem the TA said to use PFD but it clearly does not work.
 
  • #4
akbar786 said:
so when there is only dx sitting on the top i cannot use partial fraction decomposition? How can i tell if its already in expanded form. On the directions of this problem the TA said to use PFD but it clearly does not work.
If you use partial fraction decomposition, I'm reasonably certain that you'll end up with exactly what you started with. Take vela's advice.
 
  • #5
You want to use partial fractions when you have distinct factors on the bottom. For example, you would use the method if the denominator was (x+1)(x2+1). On the other hand, if you had just (x+1)2 in the denominator, it's already in the form you want it in. For instance, if you tried to expand, 1/(x+1)2, you'd write

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

But the LHS already looks just like the second term on the RHS with B=1. Similarly, in this problem, the integrand is already in the form of a possible term you'd have after doing the partial fractions expansion, which is why you're just ending up with what you started with.

Your TA probably just glanced at the problem and told you to use partial fractions before thinking about it. When you see that type of integrand, partial fractions is one of the first methods that comes to mind.
 
  • #6
oh, i see what you mean. Basically if there is only one type of polynomial squared or cubed in the denominator then i can't use partial fraction but if suppose there is (x+2)(x-1)^2 in the denominator then i could use partial fractions and expand the as a/(x+2) + b/(x-1) + c/(x-1)^2 right? Also, in order to solve the first question i posted should i complete the square for the denominator in order to get it into a trig substitution format?
 
  • #7
Yes and yes.
 
  • #8
After i complete the square i end up with dx/((x+2)^2+1)^2. I substitute u = x+2 , du = dx
so i get the integral du/(u^2+1)^2. Then i got u = tan theta, du = sec^2(theta) d(theta) and i replace that in the integral as sec^2(theta) d(theta)/(tan^2(theta) +1)^ and i get after i substitute sec^2 in the bottom for tan^2 + 1 and then cancel the sec^2 on the top i end up with 1/sec^2, which is cos^2. Am i on the right path so far?
 
  • #9
Yup!
 

What is partial fraction decomposition?

Partial fraction decomposition is a method used in calculus to decompose a rational function into simpler fractions. This is done by breaking down the denominator into its individual factors and finding the corresponding coefficients for each factor in the numerator.

Why is partial fraction decomposition useful?

Partial fraction decomposition is useful because it allows us to solve integrals of rational functions that would otherwise be difficult or impossible to solve. It also helps simplify complex expressions and make them easier to work with.

How do I perform partial fraction decomposition?

To perform partial fraction decomposition, you first need to factor the denominator of the rational function. Then, for each distinct factor, you set up a system of equations using the coefficients of the numerator and denominator. Finally, you solve the system of equations to find the coefficients of the partial fractions.

What are the different types of partial fraction decomposition?

There are two types of partial fraction decomposition: proper and improper. Proper partial fraction decomposition is used when the degree of the numerator is less than the degree of the denominator. Improper partial fraction decomposition is used when the degree of the numerator is equal to or greater than the degree of the denominator.

What are some tips for solving problems involving partial fraction decomposition?

Some tips for solving problems involving partial fraction decomposition include: fully factoring the denominator, checking for common factors between the numerator and denominator, setting up and solving the system of equations carefully, and checking your final answer by simplifying the original expression.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
937
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
729
  • Calculus and Beyond Homework Help
Replies
4
Views
908
  • Calculus and Beyond Homework Help
Replies
2
Views
497
  • Calculus and Beyond Homework Help
Replies
14
Views
372
  • Calculus and Beyond Homework Help
Replies
8
Views
865
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top