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Math Help - Integral Rp

  1. #1
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    Integral Rp

    I need solutions ...

     \int_0^2 \frac{x}{(x-1)^3}dx

     \int_0^2 \frac{1}{\sqrt{x(2-x)}}dx
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  2. #2
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    Quote Originally Posted by gilyos View Post
    I need solutions
    And I need effort from you, unless that these are "proposed problems."
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by gilyos View Post
    I need solutions ...

     \int_0^2 \frac{x}{(x-1)^3}dx

     \int_0^2 \frac{1}{\sqrt{x(2-x)}}dx
    Where are you getting stuck?
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  4. #4
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    Quote Originally Posted by gilyos View Post
    I need solutions ...

     \int_0^2 \frac{x}{(x-1)^3}dx

     \int_0^2 \frac{1}{\sqrt{x(2-x)}}dx
    1. Expand the denominator and then divide each term in the denominator by x to simplify. Else use Integration by Parts with u = x and dv = (x - 1)^{-3}.


    2. Expand the denominator, Complete the Square and use a Trigonometric Substitution.
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  5. #5
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    Red face

    u can solve the secound Problem ?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by gilyos View Post
    u can solve the secound Problem ?
    x(2-x)=2x-x^2=-(x^2-2x)=-(x^2-2x-1+1)=-(x^2-2x+1)+1=1-(x-1)^2

    So can you take \int\frac{1}{\sqrt{1-(x-1)^2}}dx?
    Last edited by VonNemo19; January 30th 2010 at 09:36 PM. Reason: duuurrrrrrrrr......drool-drool. :) Thanks J.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by gilyos View Post
    u can solve the secound Problem ?
    You still have not shown any work. Try to follow some of the hints you were given.

    For the first one, I'd use substitution, u = x - 1

    For the second, substitution u = \sqrt x. you will get it in a form where you can use \int \frac {1}{\sqrt{a^2 - x^2}}~dx = \sin^{-1} \frac xa + C to finish off the problem
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