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Math Help - Intergration Help

  1. #1
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    Intergration Help

    how do i intergrate:

    int(e^x(sqrt(1+e^x))) u=1+e^x

    and

    int((x^2)(e^x))
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  2. #2
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    Quote Originally Posted by Stylis10 View Post
    how do i intergrate:

    int(e^x(sqrt(1+e^x))) u=1+e^x

    Mr F says: u = 1 + e^x => du/dx = e^x => dx = du/e^x. So the integral becomes int e^x sqrt(u) du/e^x = int sqrt(u) du.

    and

    int((x^2)(e^x))

    Mr F says: Use integration by parts TWICE. First time will leave you with an integral of the form int x e^x dx. Second time leaves you with an integral of the form int e^x dx.
    ..
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  3. #3
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    ok i understand the second 1 but am still puzzled by the first 1
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  4. #4
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    Quote Originally Posted by Stylis10 View Post
    ok i understand the second 1 but am still puzzled by the first 1
    What part puzzles you?
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  5. #5
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    the whole thing kind of

    when i work it all out they following the steps you've shown i get the final answer as (x+e^x)^-1, but that isnt right, is it?
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  6. #6
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    Quote Originally Posted by Stylis10 View Post
    the whole thing kind of

    when i work it all out they following the steps you've shown i get the final answer as (x=e^x)^-1, but that isnt right, is it?
    \int \sqrt{u}\, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C = \frac{2}{3} u\, u^{1/2} + C = \frac{2}{3} u \, \sqrt{u} + C.

    Now substitute back u = 1 + e^x.

    I have no idea what you've done. The working you've done leading to your answer would no doubt reveal your error(s).
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  7. #7
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    ahhh thank you, i see where i have made my mistake, i substituted u = 1+e^x too soon. thank you for the help
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  8. #8
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    Quote Originally Posted by Stylis10 View Post

    how do i intergrate:

    int(e^x(sqrt(1+e^x))) u=1+e^x
    A more direct way is to set u^2=1+e^x\implies2u\,du=e^x\,dx,

    \int {e^x \sqrt {1 + e^x } \,dx}  = 2\int {u^2 \,du}  = \frac{2}<br />
{3}u^3  + k.

    The rest follows.
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