Integration using partial fractions?

In summary, the conversation discusses finding the integral of (1 + 2x)/(1+x^2) and the attempts made to solve it. The hint provided suggests splitting the fraction into two parts, 1/(1+x^2) and 2x/(1+x^2), to make it easier to solve.
  • #1
ombudsmansect
29
0

Homework Statement



find the integral of (1 + 2x)/(1+x^2)

Homework Equations





The Attempt at a Solution



I honestly have no idea how to solve this and have spent ages trying to figure it out. I would automatically assume using partial fractions to solve but the denominator presents a problem there were it cannot take a form of a partial fraction. substitution does not work either. So if anyone has worked with a problem type like this before your advice would b greatly appreciated even a hint woud b awesome :)
 
Physics news on Phys.org
  • #2
ombudsmansect said:

Homework Statement



find the integral of (1 + 2x)/(1+x^2)

Homework Equations





The Attempt at a Solution



I honestly have no idea how to solve this and have spent ages trying to figure it out. I would automatically assume using partial fractions to solve but the denominator presents a problem there were it cannot take a form of a partial fraction. substitution does not work either. So if anyone has worked with a problem type like this before your advice would b greatly appreciated even a hint woud b awesome :)

I seem to remember that if you have [tex]\frac{1+2x}{1+x^2}[/tex]

That its possible split this up

thusly getting [tex]\frac{1}{1+x^2} + \frac{2x}{1+x^2}[/tex]

That fact should be userable here :)
 
Last edited:
  • #3
thanks suz! i will try n crack this thing now :D
 

1. What is integration using partial fractions?

Integration using partial fractions is a method used to solve integrals that involve rational functions. It involves breaking down the rational function into simpler fractions, integrating each fraction separately, and then combining the results to find the final solution.

2. When is integration using partial fractions used?

Integration using partial fractions is commonly used when the integral involves a rational function that cannot be solved using basic integration techniques such as substitution or integration by parts. It is also used when the integral involves a fraction with a quadratic or higher degree polynomial in the denominator.

3. How do you determine the partial fractions for a given rational function?

To determine the partial fractions, the given rational function must first be factored into its irreducible factors. Then, the coefficients of the partial fractions can be found by equating the numerators and denominators of the original rational function with the partial fractions. A system of equations can be set up and solved to find the coefficients.

4. What is the general approach for solving integrals using partial fractions?

The general approach for solving integrals using partial fractions involves the following steps:

  • 1. Factor the rational function into its irreducible factors.
  • 2. Set up a system of equations using the coefficients of the partial fractions.
  • 3. Solve the system of equations to find the coefficients.
  • 4. Substitute the coefficients back into the partial fractions and integrate each fraction separately.
  • 5. Combine the results to find the final solution.

5. Are there any special cases to consider when using integration with partial fractions?

Yes, there are a few special cases to consider when using integration with partial fractions:

  • 1. If the degree of the numerator is equal to or higher than the degree of the denominator, long division may be necessary before using partial fractions.
  • 2. If the denominator contains repeated factors, the partial fractions may need to include terms with powers greater than 1.
  • 3. If the integral involves complex numbers, the partial fractions will also involve complex numbers.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
888
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
873
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
855
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Back
Top