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Thread: help! integrals!

  1. #1
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    help! integrals! - URGENT

    integral of:
    3(x^4/e^x^5)dx

    I'm pretty sure its substitution rule but I don't know what to put as the u

    u=x^4?
    u=e^x^5 help!

    and:

    integral of:
    5e^s/1+3e^s
    Last edited by s0urgrapes; Mar 6th 2008 at 12:29 AM.
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  2. #2
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    $\displaystyle {\text{Let }}u = e^{x^5 }\text{ ; }
    du = 5x^4 e^{x^5 }\text{ } dx $


    $\displaystyle \int {3\frac{{x^4 }}
    {{e^{x^5 } }}} {\text{ }}dx = \frac{3}
    {5}\int {\frac{{du}}
    {{u^2 }}} = \frac{3}
    {5}\left( { - \frac{1}
    {u}} \right) = - \frac{3}
    {{5u}} = - \frac{3}
    {{5e^{x^5 } }}
    = \boxed{ - \frac{{3e^{ - x^5 } }}
    {5}}
    $


    $\displaystyle {\text{Let }}u =1 + 3e^{s}\text{ ; }
    du = 3e^{s}\text{ } ds $


    $\displaystyle \int {\frac{{5e^s }}
    {{1 + 3e^s }}} {\text{ }}ds = \frac{5}
    {3}\int {\frac{{du}}
    {u}} = \frac{5}
    {3}\ln \left( {\left| u \right|} \right) = \frac{5}
    {3}\ln \left( {\left| {1 + 3e^s } \right|} \right) = \frac{5}
    {3}\ln \left( {1 + 3e^s } \right)

    $
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  3. #3
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    Quote Originally Posted by xifentoozlerix View Post
    $\displaystyle {\text{Let }}u = e^{x^5 }\text{ ; }
    du = 5x^4 e^{x^5 }\text{ } dx $


    $\displaystyle \int {3\frac{{x^4 }}
    {{e^{x^5 } }}} {\text{ }}dx = \frac{3}
    {5}\int {\frac{{du}}
    {{u^2 }}} = \frac{3}
    {5}\left( { - \frac{1}
    {u}} \right) = - \frac{3}
    {{5u}} = - \frac{3}
    {{5e^{x^5 } }}
    = \boxed{ - \frac{{3e^{ - x^5 } }}
    {5}}
    $


    $\displaystyle {\text{Let }}u =1 + 3e^{s}\text{ ; }
    du = 3e^{s}\text{ } ds $


    $\displaystyle \int {\frac{{5e^s }}
    {{1 + 3e^s }}} {\text{ }}ds = \frac{5}
    {3}\int {\frac{{du}}
    {u}} = \frac{5}
    {3}\ln \left( {\left| u \right|} \right) = \frac{5}
    {3}\ln \left( {\left| {1 + 3e^s } \right|} \right) = \frac{5}
    {3}\ln \left( {1 + 3e^s } \right)

    $
    thanks! just a question, how do you get du/u^2 in the first one, how come it isn't du/u
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  4. #4
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    Well, let's see what happens if you do $\displaystyle \int {\frac{{du}}{u}}$:


    $\displaystyle \int {\frac{{du}}
    {u}} = \int {\frac{{5x^4 e^{x^5 } {\text{ }}dx}}
    {{e^{x^5 } }}} = \int {5x^4 dx}
    $

    As you can see, the term $\displaystyle {e^{x^5 } }$ cancels and you end up with a different integral. With $\displaystyle {u^2 }$ however, the top $\displaystyle {e^{x^5 } }$ in the numerator is taken care of, and it leaves the $\displaystyle {e^{x^5 } }$ in the denominator as the original problem had:


    $\displaystyle \int {\frac{{du}}
    {{u^2 }}} = \int {\frac{{5x^4 e^{x^5 } {\text{ }}dx}}
    {{\left( {e^{x^5 } } \right)^2 }}} = \int {\frac{{5x^4 dx}}
    {{e^{x^5 } }}} $
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