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Homework Help: How to find the integral of x^3 e^(x^2) / (x^2 + 1)^2 ?

  1. Jun 7, 2008 #1
    Ok wow, I can't solve this :(

    [tex]\int\frac{x^3e^{x^2}}{(x^2+1)^2}dx[/tex]

    I have let u equal almost everything possible.

    [tex]u=x^2[/tex]
    [tex]d=2xdx[/tex]

    [tex]u=e^{x^2}[/tex]
    [tex]\ln u=x^2[/tex]
    [tex]\frac{du}{u}=2xdx[/tex]

    [tex]u=x^2+1 \leftrightarrow u-1=x^2[/tex]
    [tex]\ln u=\ln{(x^2+1)}[/tex]
    [tex]\frac{du}{u}=\frac{2xdx}{x^2+1}[/tex]
     
  2. jcsd
  3. Jun 7, 2008 #2
    rocomath, is the integral given as a definite integral? It is possible to find the value of the definite integral by using integration by parts (think about the change of variables a bit). I don't really see an easy u-substitution that's going to help you out, but maybe I'm just blind.
     
  4. Jun 7, 2008 #3
    OH, lol. I forgot to include what dV was ...

    But yes, I've been trying parts on it for a while now ... still nothing!
     
  5. Jun 7, 2008 #4

    tiny-tim

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    Hi rocomath! :smile:

    Go with [tex]u=x^2[/tex] again …

    you should get [tex]\int\frac{ue^u}{(1 + u)^2}du[/tex]

    Then use partial fractions, and try the obvious. :smile:
     
  6. Jun 7, 2008 #5
    Uh!!! I didn't even think of using Partial Fractions bc the people in the group said they weren't doing that yet ... so frustrated, lol.

    tiny-tim you're a life saver!
     
  7. Jun 7, 2008 #6
    Ok ... so I have

    [tex]\frac 1 2\int\frac{te^t}{(t+1)^2}dt[/tex]

    Now using Partial Fractions and leaving the constant 1/2 out for now:

    [tex]\frac{te^t}{(t+1)^2}=\frac{A}{t+1}+\frac{B}{(t+1)^2}[/tex]

    [tex]te^t=A(t+1)+B[/tex]

    if t = -1

    [tex]B=-\frac 1 e[/tex]

    I'm having a hard time solving for A ...

    I know what B is, so if t = 0

    [tex]A=-\frac 1 e[/tex] ???

    ... hmm, definitely not right!

    I've never done a problem that had a product on the left side.
     
    Last edited: Jun 7, 2008
  8. Jun 7, 2008 #7
    The method of partial fractions won't work like it normally does here since you can't find integers [itex]A[/itex] and [itex]B[/itex] such that

    [tex]t e^t = A (t + 1) + B[/tex]

    You can, however, solve for [itex]A[/itex] and [itex]B[/itex] in terms of your variable [itex]t[/itex], but it doesn't really make the problem any easier. We might find

    [tex]B = - e^{-1}[/tex]

    And so plugging that in to the equation above we find

    [tex]A = \frac{t e^t - e^{-1}}{t + 1}[/tex]

    I still suggest trying the method of integration by parts. Hint: can you get the entire denominator [itex](x^2+1)^2[/itex] into [itex]dv[/itex] such that you can easily integrate it?
     
  9. Jun 7, 2008 #8
    Ok, got it!

    Uh, sucha killer. Thanks for the hint Acetv, whewww!!!
     
  10. Jun 8, 2008 #9

    Gib Z

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    We still could have used ordinary partial fractions - I'm not sure why you thought it was even possible to find constants A and B for te^t = A(t+1) + B, but tiny-tims suggestion was to use partial fractions on what you've been taught to use it on - Rational functions.

    IE Only the [tex]\frac{t}{(t+1)^2}[/tex]. After that, multiply each term by e^t.

    You should have got [tex] \frac{e^t}{t+1} - \frac{e^{t}}{(t+1)^2} = \frac{1}{e} \left( \frac{e^u}{u} - \frac{e^u}{u^2} \right) [/tex] where u = t+1, which are simple to integrate.
     
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