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Math Help - Substitution Indefinite Integral Problem- Just need work checked

  1. #1
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    Substitution Indefinite Integral Problem- Just need work checked

    Not sure if i've done this problem right.

    \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx

    u=2x^3+7

    du=6x^2

    \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx

    \int (\frac{du}{u^{3/2}})

    \int (2u^{-3/2}+c

    2(2x^3+7)^{-3/2}+c

    (4x^3+14)^{-3/2}+c
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  2. #2
    Junior Member bebrave's Avatar
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    Yeah you are right i think. Because i solved and i found same
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  3. #3
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    Quote Originally Posted by Jim Marnell View Post
    Not sure if i've done this problem right.

    \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx

    u=2x^3+7

    du=6x^2 \, {\color{red}dx} Mr F says: Note the red stuff.

    \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx

    \int (\frac{du}{u^{3/2}})

    \int (2u^{-3/2}+c Mr F says: Wrong from here onwards. See below.

    2(2x^3+7)^{-3/2}+c

    (4x^3+14)^{-3/2}+c
    \int\! \frac{du}{u^{3/2}} = \int u^{-3/2} \, du = -2 u^{-1/2} + C etc.

    Quote Originally Posted by bebrave View Post
    Yeah you are right i think. Because i solved and i found same
    Then you're wrong too.
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  4. #4
    Junior Member bebrave's Avatar
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    Thank you Mr.Fantastic...
    Yeah you are right i think. Because i solved and i found same
    Last edited by mr fantastic; April 11th 2009 at 12:07 AM.
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  5. #5
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    Exclamation

    so the answer instead would be:

    (-4x^3-14)^{-1/2}+c
    Last edited by Jim Marnell; April 11th 2009 at 06:27 AM.
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  6. #6
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    Quote Originally Posted by Jim Marnell View Post
    so the answer instead would be:

    (-4x^3-14)^{-1/2}+c
    No.

    When you substitute u = 2x^3 + 7 back into -2 u^{-1/2} + C you do NOT get (-4x^3-14)^{-1/2}+C. The -2 does NOT and CANNOT get taken inside the brackets.

    By the way, even if the answer was 2 u^{-1/2} + C, this is NOT equal to (4x^3+14)^{-1/2}+C.
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