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Math Help - Integration help

  1. #1
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    Integration help

    can i start by rewriting it as

    = (x^-1/2)(1 +√x)^-2??
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  2. #2
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    Re: Integration help

    substitution ...

    u = 1 + \sqrt{x}

    du = \frac{1}{2\sqrt{x}} \, dx
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    Re: Integration help

    so when i derive 1 + √x does the square root turn into a negative ^-1/2 or will it be positive??
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    Re: Integration help

    Quote Originally Posted by asilvester635 View Post
    so when i derive 1 + √x does the square root turn into a negative ^-1/2 or will it be positive??
    look at my response again ...


    btw, it's differentiate, not "derive".
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  5. #5
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    Re: Integration help

    Quote Originally Posted by asilvester635 View Post
    so when i derive 1 + √x does the square root turn into a negative ^-1/2 or will it be positive??

    What is the derivative of -2\left(1+\sqrt{x}\right)^{-1}~?
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  6. #6
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    Re: Integration help

    Quote Originally Posted by asilvester635 View Post
    can i start by rewriting it as

    = (x^-1/2)(1 +√x)^-2??
    \text{Let } u = 1 + \sqrt{x} \text{ then } \frac{du}{dx} = \frac{1}{2\sqrt{x}}

    And so 2du = \frac{1}{\sqrt{x}} dx

    By plugging in u = 1 + \sqrt{x}$ and $\frac{1}{\sqrt{x}} dx = 2 du into \int_1^9 \frac{1}{\sqrt{x}(1+\sqrt{x})^2} dx we get:

    \begin{align*}\int_1^9 \frac{1}{\sqrt{x}(1+\sqrt{x})^2} dx =& \int_1^9 \frac{2}{u^2} du\\ =& \frac{-2}{u}\Big{]}_1^9 \\ =& \frac{-2}{1+\sqrt{x}}\Big{]}_1^9 \\ =& \frac{-2}{1+3} + \frac{2}{1+\sqrt{1}} \\ =& \frac{-2}{4} + \frac{2}{2} \\ =& \frac{-1}{2}+1 \\ =& \frac{1}{2}\end{align*}


    \therefore \int_1^9 \frac{1}{\sqrt{x}(1+\sqrt{x})^2} dx = \frac{1}{2}

    Check:

    integration 1/(((x)^(1/2))((1+((x)^(1/2))))^2) dx from x=1 to 9 - Wolfram|Alpha
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