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Math Help - Integration help

  1. #1
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    Integration help

    I found some integral problems online and have been using them to practice. I'm trying to integrate x(x + 1)^-5 with intervals of infinity and zero.

    The solution here given seems to take advantage of a shortcut. That integral is changed to the integral of y^-4 dy minus the integral of y^-5 dy, both with boundaries of infinity and 1, instead of infinity and zero.

    When I check the integral at Wolfram Alpha I get the integral as being (4x + 1) / 12 (x + 1)^4

    Which ties out to the same answer using the shortcut of the y^-4 minus y^-5 with the changed interval.

    Can someone walk through integrating x(x + 1)^-5 to get to the (4x+1) / 12(x+1)^4,

    and then also explain why the shortcut works of taking y^-4 - y^-5?

    I've attached the solution file, which shows the shortcut.

    Have a nice day!

    Chris
    Attached Thumbnails Attached Thumbnails Integration help-370integralproblemssol.pdf  
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  2. #2
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    Because if you use the change of variable

    y = 1 + x

    then x = y - 1 and dy = dx.


    Thus

    \int{x(1 + x)^{-5}\,dx} = \int{(y - 1)y^{-5}\,dy}

    = \int{y^{-4} - y^{-5}\,dy} after expanding the brackets.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Because if you use the change of variable...
    Here's an image, until LaTeX comes back on.



    By plugging the following code into this page (mostly done with search-and-replace)

    Code:
    \\
    \text{Because if you use the change of variable}
    \\
    \\y = 1 + x
    \\
    \\\text{then }x = y - 1\text{ and }dy = dx\text{.}
    \\
    \\
    \\\text{Thus}
    \\
    \\\int{x(1 + x)^{-5}\,dx} = \int{(y - 1)y^{-5}\,dy}
    \\
    \\ = \int{y^{-4} - y^5\,dy}\text{ after expanding the brackets.}
    Attached Thumbnails Attached Thumbnails Integration help-int_help.png  
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  4. #4
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    Thank you both - that is very helpful!
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