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Integration using various techniques

  1. Apr 20, 2014 #1
    1. $$\int e^{x+e^x}\,dx$$

    2. Relevant equations

    Substitution, integration by parts

    3. The attempt at a solution

    $$u=e^x$$ $$\int e^{x+e^x}\,dx = \int e^x e^{e^x}\,dx = \int ue^u\,du$$
    $$a=u$$ $$da=1du$$ $$dv=e^udu$$ $$v=e^u$$
    $$=ue^u-\int e^u\,du = ue^u-e^u$$ $$=e^x e^{e^x}+e^{e^x} = e^{x+e^x}-e^{e^x} + C$$

    The textbook's answer was $$e^{e^x} +C$$ How come my answer is different? Any help is much appreciated.
     
  2. jcsd
  3. Apr 20, 2014 #2

    Mark44

    Staff: Mentor

    What is du? It is very important to also write du when you make a substitution u = <something>. In this case (and often), du is NOT equal to dx.
     
  4. Apr 20, 2014 #3
    $$du=e^xdx$$
    How does this work into the calculation?
     
  5. Apr 20, 2014 #4

    haruspex

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    But that's not what you did. You turned ##e^xdx## into ##udu##.
     
  6. Apr 20, 2014 #5
    ∫ex+exdx=∫exeexdx=∫ueudu

    Do you mean I should do something like this? $$\int e^x\,dx = \int e^xe^{e^x}\,dx = \int ue^u\,e^x dx$$ But the point of my substitution was to turn everything into u's...
     
  7. Apr 20, 2014 #6

    Mark44

    Staff: Mentor

    You have ##\int e^x e^{e^x}dx##
    u = ex, so du = exdx

    What does your integral look like using this substution?
     
  8. Apr 20, 2014 #7
    Oh, I got it now. Thanks! My integral would be

    $$\int e^u\,du$$
     
  9. Apr 20, 2014 #8

    Mark44

    Staff: Mentor

    Yes.
     
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