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Math Help - integral using substitution

  1. #1
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    integral using substitution

    I'm not sure what I did wrong here..
     \int 60x \sqrt {1+x} dx
     u= \sqrt {1+x}
     x= u^2-1
     dx= 2u du

    Then I got

     60 \int (u^3-u) 2u du

    After I integrated I got

     120 (\frac {1} {5} u^5- \frac {1} {3} u^3)
    which isn't one of the answer choices...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Linnus View Post
    I'm not sure what I did wrong here..
     \int 60x \sqrt {1+x} dx
     u= \sqrt {1+x}
     x= u^2-1
     dx= 2u du

    Then I got

     60 \int (u^3-u) 2u du

    After I integrated I got

     120 (\frac {1} {5} u^5- \frac {1} {3} u^3)
    which isn't one of the answer choices...
    maybe because your answer is in terms of u when it should be in terms of x...
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  3. #3
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    hmm the answer are all in terms of u
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Linnus View Post
    hmm the answer are all in terms of u
    it's possible that they did a different substitution from you did (u = x + 1 could work, for instance), or you need to simplify your answer to get one of the others, or your answer and the correct choice differ by a constant

    what are the answer choices?
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  5. #5
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    The closest answer choice is  25u^5-40U^3+C ...i wonder if this is a misprint or not. The question told me to used  u= \sqrt {1+x}
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Linnus View Post
    The closest answer choice is  25u^5-40U^3+C ...i wonder if this is a misprint or not. The question told me to used  u= \sqrt {1+x}
    yes, it's a misprint, it should be 24u^5
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