Integration by parts expression help

In summary, the expression to integrate is \int x^{3}e^{x^{2}}dx and the recommended approach is to set u=x^{3} and dv=e^{x^{2}}dx. However, this leads to a difficult integral and a u-substitution does not work. The alternative approach of setting u=e^{x^{2}} and dv=x^{3}dx and using integration by parts also leads to a challenging integral. Further advice is needed.
  • #1
AStaunton
105
1
the expression to integrate is:

[tex]\int x^{3}e^{x^{2}}dx[/tex]

and in the spirit of "LIATE" I set my u and dv as the following:

[tex]dv=e^{x^{2}}dx[/tex]

[tex]u=x^{3}[/tex]

however, doing this that I integrate [tex]dv=e^{x^{2}}dx[/tex] in order to get v...and unless I'm missing something, this does not seem like an easy integral! a u substition won't work as i'd need an x^1 term multiplying by the e^x^2...

and going the other way and setting dv=x^3dx and u=e^x^2 and plugging into int,by parts formula gets:

[tex]\frac{x^{4}}{4}e^{x^{2}}-\int\frac{x^{4}}{4}2xe^{x^{2}}dx[/tex]

and I don't think further integration by parts will help with this new integral..

any advice appreciated.
 
Physics news on Phys.org
  • #2
correction of typo:



however, doing this REQUIRES that I integrate [tex]dv=e^{x^{2}}dx[/tex] in order to get v...
 
  • #3
change variable x^2->u
 
  • #4
Perform a u-substitution with u = x3, and then integrate by parts.
 

Related to Integration by parts expression help

1. What is integration by parts?

Integration by parts is a method used in calculus to find the integral of a product of two functions. It is based on the product rule for differentiation and is useful for integrating functions that cannot be easily solved using other methods.

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

You can use integration by parts when the integral you are trying to solve is a product of two functions. This method is also helpful when one of the functions in the product is easier to integrate after a few iterations of the integration by parts formula.

3. What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du. This formula uses the product rule for differentiation to solve integrals of products of two functions.

4. Can you provide an example of using integration by parts?

Sure, let's say we want to solve the integral of x sin(x). We can use integration by parts by letting u = x and dv = sin(x) dx. This gives us du = dx and v = -cos(x). Plugging these values into the formula, we get ∫x sin(x) dx = -x cos(x) - ∫-cos(x) dx. This simplifies to ∫x sin(x) dx = -x cos(x) + sin(x) + C, where C is a constant of integration.

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

Yes, there are a few tips that can make integration by parts easier. First, try to choose u and dv so that the new integral becomes simpler after one or two iterations of the formula. Also, if the integral you are trying to solve involves trigonometric functions, try using trigonometric identities to simplify the expression before using integration by parts.

Similar threads

  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
812
  • Calculus and Beyond Homework Help
Replies
15
Views
831
  • Calculus and Beyond Homework Help
Replies
7
Views
781
  • Calculus and Beyond Homework Help
Replies
14
Views
444
  • Calculus and Beyond Homework Help
Replies
3
Views
403
  • Calculus and Beyond Homework Help
Replies
1
Views
795
  • Calculus and Beyond Homework Help
Replies
7
Views
979
  • Calculus and Beyond Homework Help
Replies
10
Views
567
  • Calculus and Beyond Homework Help
Replies
2
Views
339
Back
Top