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U-substitution problem

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations


    3. The attempt at a solution

    First, I let ##u = x + 1##. Then ##du = dx## and ##x = u - 1##.

    So ##\displaystyle \int \frac{u - 1}{\sqrt{u}}du = \int u^{\frac{1}{2}} - u^{-\frac{1}{2}}du = \frac{2}{3}u^{\frac{3}{4}} - 2u^{\frac{1}{2}} + C = \frac{2}{3}u\sqrt{u} - 2u + C = \frac{2}{3}u(\sqrt{u} - 3) + C = \frac{2}{3}(x + 1)(\sqrt{x + 1} - 3) + C##

    However, the answer book says that the correct answer is ##\displaystyle \frac{2}{3}(x - 1)\sqrt{x + 1} + C##

    What am I doing wrong?
     
  2. jcsd
  3. Dec 8, 2015 #2

    Samy_A

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    Check what ##\int u^{p}du## is for ##p \neq -1##. You made a mistake when computing one of the integrals.

    Also, in the above step something went wrong.

    Hmm, I find a slightly different result than what your book says ...
     
    Last edited: Dec 8, 2015
  4. Dec 8, 2015 #3

    LCKurtz

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    You don't mean ##u^{\frac{3}{4}}##there.
    and you don't mean ##2u## there.

    Fix that, and with a little care you will find the text answer is correct.
     
    Last edited: Dec 8, 2015
  5. Dec 8, 2015 #4

    Ray Vickson

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    The book's answer is incorrect; differentiate it wrt x to see that it fails to give back the integrand ##x/\sqrt{x+1}##.
     
  6. Dec 9, 2015 #5

    LCKurtz

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    Yes. I made the same silly error the book did. The text answer should have been$$
    \frac{2}{3}(x - 2)\sqrt{x + 1} + C$$
     
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