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Issues with u-Substitution

  1. Jul 2, 2012 #1
    1. The problem statement, all variables and given/known data
    I am struggling with u-substitution. I understand that it essentially the undoing of the chain rule, but I don not get how to actually go about the procedure.

    I have this example from my textbook:
    [itex]\int[/itex]4x[itex]\sqrt{x^{2}+1}[/itex]dx

    It says that u=x^2+1 and that du/2=xdx. Where did they get this from? How do I know what to use for u in any given equation. Once I have these to values (u and du) I know that it is just a matter of taking the integral andd then plugging x in for u. Any help is appreciated.
     
    Last edited by a moderator: Jul 2, 2012
  2. jcsd
  3. Jul 2, 2012 #2

    Mark44

    Staff: Mentor

    Do you know how to get the differential of u (i.e., du)?

     
  4. Jul 2, 2012 #3
    If I tske the derivative of u, then that should give me du/dx=something. I then multiply by dx to get du. Is that what you mean?
     
  5. Jul 2, 2012 #4
    That is right, you can treat it just like a fraction. Then isolate dx and substitute it into the original equation along with your U substitution. You'll lean what to substitute U for with experience. Basically, what ever you substitute U for, call it A for you'll end up with [itex]\frac{1}{\frac{d}{dx} A}[/itex]. If you do it right, this factor will cancel with some other x in the integral, leaving you with an integral only in terms of U du.
     
    Last edited: Jul 2, 2012
  6. Jul 2, 2012 #5
    Thank you for the replies. I'm sorry, but I don't understand anything about A. I have never heard of this before. Could you elaborate please?

    Also, are there any tricks to figuring out what u is?
     
  7. Jul 2, 2012 #6
    That is the "trick". I said let A represent whatever you substitute for U. Then when you follow the method you'll end up with 1 over A prime. Will post an example, just a sec.....
     
  8. Jul 2, 2012 #7

    SammyS

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    So, if [itex]\displaystyle u=x^2+1\,,[/itex] then [itex]\displaystyle \frac{du}{dx}=2x\,.[/itex]

    Therefore, [itex]\displaystyle du=2x\,dx\,,[/itex] correct ?

    Now, divide by 2.
     
  9. Jul 2, 2012 #8
    subsitiution.jpg

    I forgot to write dx at the end of original function at top of page, I hope it's still clear.
     
    Last edited: Jul 2, 2012
  10. Jul 2, 2012 #9
    ^That right there just solved my problems. Thank you so much for your help everyone I really appreciate it.

    One more question though. How do I go about solving u-substitution problems for definite integrals? Is it the same but I plug in values of x and subtract at the end?
     
  11. Jul 2, 2012 #10
  12. Jul 4, 2012 #11
    One more thing, are there any tricks to realizing if a problem is u-substitution or integration by parts? Or is it merely just a matter of doing one to see if it works?
     
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