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Integrate: e(x^2 +x)(2x+1) dx

  1. Jul 1, 2009 #1

    a.a

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    1. The problem statement, all variables and given/known data

    integrate: e(x^2 +x)(2x+1) dx


    3. The attempt at a solution

    let u= e(x^2 +x)
    du=e(x^2 +x)(2x+1)dx

    integral e(x^2 +x)(2x+1) dx = integral 1/u du

    am I on the right track? i didnt get the same answer as the prof...
     
  2. jcsd
  3. Jul 1, 2009 #2

    statdad

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    Homework Helper

    Re: integration

    If you mean

    [tex]
    \int e^{x^2+x} (2x+1) \, dx
    [/tex]

    then your substitution is one of at least two that will work, but how do you obtain

    [tex]
    \int \frac 1 u \, du
    [/tex]

    as the next step?
     
  4. Jul 1, 2009 #3

    a.a

    User Avatar

    Re: integration

    because du=e(x2 +x)(2x+1)dx ?

    integral e(x2 +x)(2x+1) dx = integral du/e(x2 +x)
    =integral 1/u du
    ?
     
  5. Jul 1, 2009 #4

    statdad

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    Homework Helper

    Re: integration

    If

    [tex]
    u = e^{x^2+x}
    [/tex]

    then

    [tex]
    du = e^{x^2+x}(2x+1) dx
    [/tex]


    which is exactly the form of the original integral.
    why is there need for a fraction?
     
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