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Math Help - Integrate a fraction with limits, with the route of x on the bottom

  1. #1
    Senior Member Mukilab's Avatar
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    Integrate a fraction with limits, with the route of x on the bottom

    I have no idea how to do this:

    Evaluate the following integral::

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

    I thought the bottom can be separated into x^-.05 and 1 which wouldn't be too bad to go from there but I don't think that's the answer.

    As soon as I can get x to a power I think I can integrate it into ln(something) but I'm not sure how to do that first step.
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  2. #2
    MHF Contributor

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    Re: Integrate a fraction with limits, with the route of x on the bottom

    Quote Originally Posted by Mukilab View Post
    I have no idea how to do this:
    Evaluate the following integral::
    \int \frac {1}{\sqrt{x+1}} dx
    What is the derivative of 2\sqrt{x+1}~?
    Thanks from Mukilab and topsquark
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  3. #3
    Senior Member Mukilab's Avatar
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    Re: Integrate a fraction with limits, with the route of x on the bottom

    oh like chain rule?

    So the derivative of that is

    2(0.5)(1)(x+1)^{-0.5} = (x+1)^{-0.5}

    Huh. How did you figure out it's 2sqrt(x+1) though? Are there any steps or was it just intuition?
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  4. #4
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    Re: Integrate a fraction with limits, with the route of x on the bottom

    Quote Originally Posted by Mukilab View Post
    I have no idea how to do this:

    Evaluate the following integral::

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

    I thought the bottom can be separated into x^-.05 and 1 which wouldn't be too bad to go from there but I don't think that's the answer.

    As soon as I can get x to a power I think I can integrate it into ln(something) but I'm not sure how to do that first step.
    \displaystyle \begin{align*} \int{ \frac{1}{\sqrt{x + 1}}\,dx} &= \int{ \left( x + 1 \right) ^{-\frac{1}{2}}\,dx} \end{align*}

    If you're good with your integration processes, you can get the integral from here, otherwise make a substitution \displaystyle \begin{align*} u = x + 1 \implies du = dx \end{align*}.
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  5. #5
    Senior Member Mukilab's Avatar
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    Re: Integrate a fraction with limits, with the route of x on the bottom

    Thanks.

    I have no idea how to integrate something in brackets, so I've gone with the u thing.

    So  \int (x+1)^\frac{-1}{2} dx = \int u^\frac{-1}{2} du = [2u^\frac{1}{2}] + c = 2(x+1)^\frac{1}{2} + c = 2 \sqrt{(x+1)} + c is that correct?
    Last edited by Mukilab; October 22nd 2013 at 06:11 AM.
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  6. #6
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    Re: Integrate a fraction with limits, with the route of x on the bottom

    Yes that's correct, well done
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