U substitution and integration by parts

  • #1
I would think because of this

Screenshot2012-05-23at73955PM.png


The following problem:

Screenshot2012-05-23at72700PM.png


At this stage they should use integration by parts:

Screenshot2012-05-23at74154PM.png


However, maybe integration by parts is only useful when one of the parts is e^x ln or a trigonometric formula.
 
Last edited:

Answers and Replies

  • #2
94
1
Integration by parts can be useful whenever the integrand is a product of two functions. But it is not always the easiest method to use. For instance, the integral [itex]\int(u-1)\sqrt{u}du=\int(u^{3/2}-u^{1/2})du[/itex] can easily be solved using the power rule in reverse. Of course, you could solve it using integration by parts as well, but it's just more work than is necessary.
 
  • #3
good, thanks.
 
  • #4
294
0
What would be your u and what would be your dv?

Using integration by parts might work, but I feel that even if it does work it will be much more work than using u substitution.


Is the problem here that you don't find the u substitution they used to be 'legal'?
 
  • #5
Yea, it doesn't seem legal, because I thought you couldn't take the product of two functions in an integrand.
 
  • #6
jgens
Gold Member
1,583
50
Yea, it doesn't seem legal, because I thought you couldn't take the product of two functions in an integrand.

You need to work on being more precise. The integrand of [itex]\int_0^1 x^2 \mathrm{d}x[/itex] is the product of two functions but is clearly integrable.
 

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