Algebraic Manipulation within Integration by Parts problem

In summary, the conversation is about how to solve an integration by parts problem involving the expression \int\frac{x^{3}}{1+x^{2}}dx. The expert suggests using polynomial long division or substitution to simplify the problem. The steps for polynomial long division are explained and it is mentioned that it is similar to division with numbers. The conversation ends with the expert giving tips on how to solve the problem using long division.
  • #1
Lebombo
144
0

Homework Statement



I'm doing an integration by parts problem. After setting up [∫udv = uv - ∫vdu], I come across this situation. How is this algebraic manipulation accomplished?

From: [itex]\int\frac{x^{3}}{1+x^{2}}dx[/itex]

To: [itex]\int x - \frac{x}{1+x^{2}}dx[/itex]
 
Physics news on Phys.org
  • #2
By polynomial long division, for example.
Alternatively, write x^3=x*(x^2+1-1), and see how that might help you.
 
  • #3
I dread polynomial long division because I never had the opportunity to learn it in school. Skipped most of college algebra due to military and never took precalc, which has created a lot of struggling spots in Calculus How do I even set this up for long division? I gave it a try in the attached picture. Is this how it should be set up? Or should I rewrite the numerator portion from [1+x^2] to [x^2 + x^1 + 1] as done with the denominator?

polynomiallongdivision_zps96bbb1fc.png
 
Last edited:
  • #4
Strangely, it is exactly the same as elementary school long division.

Set your divisor up as ##x^2 + 1## instead of ## 1 + x^2##. Then divide ##x^2 \text{ into }x^3## getting a (partial) quotient of x. Place that on top of your division sign just as you did in 5th grade long division.

Continue with your 5th grade long division by multiplying your x by ##x^2 + 1## and placing the product below the ##x^2 + 1##. Now subtract as you would in an ordinary problem. In this case what you have is a remainder, of -x so you put that over the divisor ##x^2 + 1##.

While you do at this point have two integrals that are easy to compute, the step was not necessary. You could have done the original by substitution: u = ##x^2##. Try it.
 
  • #5
What you have started is fine. If you learned how to do long division with numbers, division with polynomials is quite similar.

Step 1: what factor must you multiply (1+x^2) by to obtain an expression with x^3 in it?

Step 2: write the factor from step 1 on top over the x^3

Step 3: multiply the factor from Step 1 by (1 + x^2) and write the result below the polynomial inside the bracket. Write like powers underneath like powers.

Step 4: Subtract the product from the original polynomial being divided.

Step 5: If the result of the subtraction is a polynomial of smaller degree than(1+x^2), you can stop. Otherwise, repeat steps 1-4 above.

The quotient will be the entire expression on top of the brackets, and the remainder will be the expression of smaller degree than the divisor which is at the bottom of all of the subtractions.

Enjoy!
 

Related to Algebraic Manipulation within Integration by Parts problem

1. What is algebraic manipulation within integration by parts?

Algebraic manipulation within integration by parts is a technique used to solve integrals involving products of functions. It involves rearranging the terms and applying the integration by parts formula to simplify the integral.

2. How do I know when to use algebraic manipulation within integration by parts?

You can use algebraic manipulation within integration by parts when the integral involves a product of functions, and the integration by parts formula does not lead to a simpler integral. In this case, rearranging the terms may help simplify the integral.

3. What are the steps for using algebraic manipulation within integration by parts?

The steps for using algebraic manipulation within integration by parts are:

  1. Rearrange the terms in the integral to have a product of functions
  2. Apply the integration by parts formula to the new integral
  3. Simplify the integral using algebraic manipulation techniques
  4. Integrate the simplified integral

4. Can algebraic manipulation within integration by parts be used for any type of integral?

No, algebraic manipulation within integration by parts is only useful for integrals with products of functions. It cannot be used for integrals that cannot be expressed as a product of functions.

5. Are there any tips for making algebraic manipulation within integration by parts easier?

One tip for making algebraic manipulation within integration by parts easier is to choose the best rearrangement of terms to simplify the integral. This may require some trial and error, but it can lead to a simpler integral and easier integration. Another tip is to carefully apply algebraic manipulation techniques, such as factoring and substitution, to simplify the integral before integrating.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
384
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
791
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
758
  • Calculus and Beyond Homework Help
Replies
8
Views
798
  • Calculus and Beyond Homework Help
Replies
5
Views
824
  • Calculus and Beyond Homework Help
Replies
10
Views
539
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Back
Top