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Thread: Getting Rid of a Constant of 1 in Integral Substitution

  1. #1
    MHF Contributor Jason76's Avatar
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    Getting Rid of a Constant of 1 in Integral Substitution

    $\displaystyle \int \dfrac{dx}{2\sqrt{x} + 2x}$. - The key here is pulling the constant out, before doing u substitution.

    $\displaystyle \int (\dfrac{1}{2})\dfrac{dx}{\sqrt{x} + x}$.

    $\displaystyle \int (\dfrac{1}{2})\dfrac{du}{u}$.

    $\displaystyle u = \sqrt{x} + x$.

    $\displaystyle du = \dfrac{1}{2\sqrt{x}} + 1dx $. ?? What is the next step? Somehow we need to get rid of that constant of 1, and $\displaystyle \dfrac{1}{x}$

    Note: The answer to this problem is $\displaystyle \ln(\sqrt{x} + x) + C$ But do we get there?
    Last edited by Jason76; Sep 3rd 2014 at 03:29 AM.
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    Forum Admin topsquark's Avatar
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    Re: Getting Rid of a Constant of 1 in Integral Substitution

    Quote Originally Posted by Jason76 View Post
    $\displaystyle \int \dfrac{dx}{2\sqrt{x} + 2x}$. - The key here is pulling the constant out, before doing u substitution.

    $\displaystyle \int (\dfrac{1}{2})\dfrac{dx}{\sqrt{x} + x}$.

    $\displaystyle \int (\dfrac{1}{2})\dfrac{du}{u}$.

    $\displaystyle u = \sqrt{x} + x$.

    $\displaystyle du = \dfrac{1}{2\sqrt{x}} + 1dx $. ?? What is the next step? Somehow we need to get rid of that constant of 1, and $\displaystyle \dfrac{1}{x}$

    Note: The answer to this problem is $\displaystyle \ln(\sqrt{x} + x) + C$ But do we get there?
    That technique isn't going to work. Try the substitution $\displaystyle y = \sqrt{x}$ and see where that gets you.

    -Dan
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    Re: Getting Rid of a Constant of 1 in Integral Substitution

    Unfortunately, (it breaks my heart to tell you this), but the answer (you have) is wrong!

    $\displaystyle \int_a^b \frac{1}{2\sqrt(x) + 2x } dx = ln|\sqrt(x) +1|$

    I'll give you a critical step.

    Factor out in the denominator $\displaystyle \sqrt(x)$

    Then use the substitution $\displaystyle u = \sqrt(x)$

    See where that gets you =)
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