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Math Help - Getting Rid of a Constant of 1 in Integral Substitution

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    Getting Rid of a Constant of 1 in Integral Substitution

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

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

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

    u = \sqrt{x} + x.

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

    Note: The answer to this problem is \ln(\sqrt{x} + x) + C But do we get there?
    Last edited by Jason76; September 3rd 2014 at 04:29 AM.
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    Re: Getting Rid of a Constant of 1 in Integral Substitution

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

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

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

    u = \sqrt{x} + x.

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

    Note: The answer to this problem is \ln(\sqrt{x} + x) + C But do we get there?
    That technique isn't going to work. Try the substitution 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!

    \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  \sqrt(x)

    Then use the substitution u = \sqrt(x)

    See where that gets you =)
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