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Thread: Evaluating Indefinate Integerals

  1. #1
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    Evaluating Indefinate Integerals

    #37, I know to make u = 1 + x but I am lost after that..

    https://imgur.com/a/VZXox
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    Re: Evaluating Indefinate Integerals

    Quote Originally Posted by MrJank View Post
    #37, I know to make u = 1 + x but I am lost after that..

    https://imgur.com/a/VZXox
    If $u=1+x$ then $\dfrac{{xdx}}{{\sqrt {1 + x} }} = \dfrac{{u - 1}}{{\sqrt u }}du = \left( {\sqrt u - \dfrac{1}{{\sqrt u }}} \right)du$
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    Re: Evaluating Indefinate Integerals

    Quote Originally Posted by MrJank View Post
    #37, I know to make u = 1 + x

    https://imgur.com/a/VZXox
    No, you are not required to make u = 1 + x. Be open to other expressions.


    \displaystyle\int\dfrac{x}{\sqrt{1 + x}}dx


    For example,


    let  \ \  u \ = \ \sqrt{1 + x}.

    Then  \ \ u^2 \  = \ 1 + x.

     x \ = \ u^2 - 1

    dx \ = \ 2u \ du


    Substitute:


    \displaystyle\int \bigg(\dfrac{u^2 - 1}{u}\bigg)2u \ du

    \displaystyle\int\(2u^2 \ - \ 2)du
    Last edited by greg1313; Dec 6th 2017 at 07:30 AM.
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    Re: Evaluating Indefinate Integerals

    It is an interesting exercise to do this problem with the substitution u= x+ 1, so integrating \int u^{1/2}- u^{-1/2} du as Plato gives, then doing the problem with u= \sqrt{x+ 1}, so integrating \int 2u^2- 2 du, as greg1313 suggests, and showing that the results are the same.
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    Re: Evaluating Indefinate Integerals

    Quote Originally Posted by HallsofIvy View Post
    It is an interesting exercise to do this problem with the substitution u= x+ 1, so integrating \int u^{1/2}- u^{-1/2} du as Plato gives, then doing the problem with u= \sqrt{x+ 1}, so integrating \int 2u^2- 2 du, as greg1313 suggests, and showing that the results are the same.
    Or at least differing only by a constant :P
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