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Math Help - Integral

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Integral

    I dont know why but I am blanking on how to do this one

    \int\frac{x^{\frac{1}{4}}}{1+\sqrt{x}}dx

    I know its sad...it will be something obvious...I tam thinking trig sub?
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I dont know why but I am blanking on how to do this one

    \int\frac{x^{\frac{1}{4}}}{1+\sqrt{x}}dx

    I know its sad...it will be something obvious...I tam thinking trig sub?
    try x=u^4 \to dx=4u^3du

    \int\frac{x^{\frac{1}{4}}}{1+\sqrt{x}}dx=\int \frac{4u^4}{1+u^2}du

    by long divison

    \int \left( 4u^2-4+\frac{4}{1+u^2}\right)du

    from here it is off to the races
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    To split the last integral, it's not necessary to do a long division. (Just in case.)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    To split the last integral, it's not necessary to do a long division. (Just in case.)
    Let me see if I can tell what you are thinking

    \frac {4u^4}{1 + u^2} = \frac {4u^4 - 4 + 4}{1 + u^2} = \frac {4u^4 - 4}{1 + u^2} + \frac {4}{1 + u^2} =  \frac {4(u^2 + 1)(u^2 - 1)}{1 + u^2} + \frac 4{1 + u^2} = 4u^2 - 4 + \frac 4{1 + u^2}

    That wasn't so bad ...it was even easier knowing the EmptySet's final form to work towards...
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  5. #5
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by Jhevon View Post
    That wasn't so bad ...it was even easier knowing the EmptySet's final form to work towards...
    Of course!
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