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Math Help - Integral problem

  1. #1
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    Integral problem

    So I did this integral problem (attachment), but I didn't get the same answer as the book. Looks like the book made the substitution u= x^(1/2) +1, but I don't see how that makes the problem any easier... and I don't see were I did anything wrong.

    Any help? Thanks!
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  2. #2
    Super Member Quacky's Avatar
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    Re: Integral problem

    I tried to follow your working but I'm afraid I got lost. In these situations, it's best to just forget it and start again.

    I'm assuming you want to stick with your substitution so I'm going to try it.

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

    Letting u=\sqrt{x} gives du=\frac{1}{2\sqrt{x}} dx \Rightarrow dx=2\sqrt{x}~du \Rightarrow dx=2u~du

    The integral becomes:

    \int\frac{u^2}{u+1}\cdot 2u~du

    =2\int\frac{u^3}{u+1}~du

    =2\int\frac{u^3+1-1}{u+1}~du

    =2\int[\frac{u^3+1}{u+1}-\frac{1}{u+1}]~du

    =2\int[\frac{(u+1)(u^2-u+1)}{u+1}-\frac{1}{u+1}]~du using the sum of two cubes formula.

    =2\int{(u^2-u+1}-\frac{1}{u+1})~du

    Which you can now integrate term by term.

    Edit: I notice, upon reflection, that you already had this exact thing. See my following post.
    Last edited by Quacky; December 8th 2011 at 06:15 PM.
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  3. #3
    Super Member Quacky's Avatar
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    Re: Integral problem

    Whoops, I didn't notice that this is exactly what you had. As far as I can tell, your answer is fine. What did the book have?
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  4. #4
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    Re: Integral problem

    My book had this answer (attachment):

    Not sure if they're equivalent or not... but I still don't see any errors on my part. Thanks for helping out!
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  5. #5
    Super Member Quacky's Avatar
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    Re: Integral problem

    I highly suspect they'll be equivalent. Your working is fine.
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