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Math Help - U-substitution with integrals

  1. #1
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    U-substitution with integrals

    Hello.

    I have been working with U-substitution the last two days, and things have been fine, until I got to a certain problem. I went about it the same way that I have done with all of my other problems, and yet the answer is wrong, and I dont understand why. Rather than try to code it out here, I took a picture from my workbook, and I hope that is ok.

    Can someone explain to me why I cant substitute like I have? As far as I can see after looking over it several times, I have done things exactly as I always have. I have a gut feeling that it has to do with the fact that there is the 1 + sqrt X in the denominator, but I still dont see why it wouldnt work. Thanks!

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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: U-substitution with integrals

    I notice you bring a factor involving x outside of the integral, and you cannot do this. Using your substitution, we find:

    u=1+\sqrt{x}\,\therefore\,du=\frac{1}{2\sqrt{x}}\,  dx\,\therefore\,dx=2(u-1)\,dx

    and now we have:

    2\int\frac{u-1}{u}\,du=2\int 1-\frac{1}{u}\,du
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  3. #3
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    Re: U-substitution with integrals

    Ok, so you can only take constants outside of the integral, not variables? I wasnt aware of that... So what I should have done was kept all of it inside, cancelled out like I did, but integrated 2 sqrt x as well... Correct?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: U-substitution with integrals

    When you make a u-substitution, you don't want any x's left. You want to rewrite the integral in terms of u only.
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