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Thread: root of x in denominator and not yet been taught to integrate quotients

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    root of x in denominator and not yet been taught to integrate quotients

    Hello

    $\displaystyle \int \frac{3x+1}{\sqrt(2)}$

    And I have not yet been taught only basic integration. ie not how to do quotients. So how do I make this into a simple equation that I can integrate?

    Angus
    Last edited by angypangy; Nov 13th 2011 at 01:27 AM. Reason: spelling mistake
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    Re: root of x in denominator and not yet been taught to integrate quotients

    You can put the constant $\displaystyle \frac{1}{\sqrt{2}}$ outside the integral:
    $\displaystyle \int \frac{3x+1}{\sqrt{2}}dx=\frac{1}{\sqrt{2}}\int (3x+1)dx$

    Now you have a basic integral.
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    Re: root of x in denominator and not yet been taught to integrate quotients

    Quote Originally Posted by angypangy View Post
    Hello

    $\displaystyle \int \frac{3x+1}{\sqrt(2)}$

    And I have not yet been taught only basic integration. ie not how to do quotients. So how do I make this into a simple equation that I can integrate?

    Angus
    $\displaystyle \sqrt{2}$ is a constant and so therefore is it's reciprocal $\displaystyle \dfrac{1}{\sqrt{2}}$. If you've been taught how to rationalise the denominator then this is also $\displaystyle \dfrac{\sqrt2}{2}$ which is still a constant.

    $\displaystyle \int \dfrac{3x+1}{\sqrt{2} dx} = \dfrac{\sqrt2}{2}\int (3x+1) dx$


    Note that $\displaystyle \dfrac{\sqrt{2}}{2}C$ is also a constant (where C is your normal constant of integration)
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    Re: root of x in denominator and not yet been taught to integrate quotients

    Quote Originally Posted by angypangy View Post
    Hello

    $\displaystyle \int \frac{3x+1}{\sqrt(2)}$

    And I have not yet been taught only basic integration. ie not how to do quotients. So how do I make this into a simple equation that I can integrate?

    Angus
    I expect, from the title of the thread, that the OP meant to ask how to integrate this: $\displaystyle \displaystyle \int{\frac{3x + 1}{\sqrt{x}}\,dx}$, in which case...

    $\displaystyle \displaystyle \begin{align*} \int{\frac{3x+1}{\sqrt{x}}\,dx} &= \int{\frac{3x}{\sqrt{x}} + \frac{1}{\sqrt{x}}\,dx} \\ &= \int{3\sqrt{x} + \frac{1}{\sqrt{x}}\,dx} \\ &= \int{3x^{\frac{1}{2}} + x^{-\frac{1}{2}}\,dx} \end{align*}$

    and you should now be able to integrate using the rules you know
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    Re: root of x in denominator and not yet been taught to integrate quotients

    Quote Originally Posted by Prove It View Post
    I expect, from the title of the thread, that the OP meant to ask how to integrate this: $\displaystyle \displaystyle \int{\frac{3x + 1}{\sqrt{x}}\,dx}$, in which case...

    $\displaystyle \displaystyle \begin{align*} \int{\frac{3x+1}{\sqrt{x}}\,dx} &= \int{\frac{3x}{\sqrt{x}} + \frac{1}{\sqrt{x}}\,dx} \\ &= \int{3\sqrt{x} + \frac{1}{\sqrt{x}}\,dx} \\ &= \int{3x^{\frac{1}{2}} + x^{-\frac{1}{2}}\,dx} \end{align*}$

    and you should now be able to integrate using the rules you know
    Yes this was the question - but glad I got it wrong - because answers to my incorrect question were useful. I had a think a think just after posting and worked out what to do. But thanks for replying.
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