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Math Help - Integration by substituion and partial factions

  1. #1
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    Integration by substituion and partial factions

    Hi, I'm trying to solve this problem by substitution and partial fractions (both methods stipulated in the question).

    \int\frac{1}{\sqrt{x}(1-\sqrt{x})(2-\sqrt{x})}\ dx

    so i substitute u=\sqrt{x} so dx=2u du and form the new integral:

    \int\frac{2u}{2u^2(1-u)(2-u)}\ du but am a) not sure if this is correct and b) when I try and break this up into partial fractions i get some numerators equal to 0 :S

    Please help!
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  2. #2
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    Quote Originally Posted by flashylightsmeow View Post
    Hi, I'm trying to solve this problem by substitution and partial fractions (both methods stipulated in the question).

    \int\frac{1}{\sqrt{x}(1-\sqrt{x})(2-\sqrt{x})}\ dx

    so i substitute u=\sqrt{x} so dx=2u du and form the new integral:

    \int\frac{2u}{2u^2(1-u)(2-u)}\ du but am a) not sure if this is correct and b) when I try and break this up into partial fractions i get some numerators equal to 0 :S

    Please help!
    After simplification, you are left with this 2\int \frac{1}{(1-u)(2-u)}du , try breaking 1/(1-u)(2-u) again.
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  3. #3
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    Here is what I purpose:

    Reduce your integral to  \int \frac {1}{u(u-1)(u-2)} , then perform partial fractions.
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  4. #4
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    Ah, ok. But one thing, doesnt the simplication leave the denominator as u(1-u)(2-u)?
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  5. #5
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    Yeah, that's what I got, too.
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  6. #6
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    Yes, mathaddict miswrote. The u in the numerator cancels one u in the denominator. Now you need A, B, C, so that
    \frac{2}{u(u-1)(u-2)}= \frac{A}{u}+ \frac{B}{u-1}+ \frac{C}u-2}. That should be easy.
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  7. #7
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    Quote Originally Posted by tttcomrader View Post
    Here is what I purpose:

    Reduce your integral to  \int \frac {1}{u(u-1)(u-2)} , then perform partial fractions.
    This does not work.

    Notice that if you let \displaystyle u = \sqrt{x} then \displaystyle du = \frac{1}{2\sqrt{x}}\,dx.

    So \displaystyle \int{\frac{1}{\sqrt{x}(1 - \sqrt{x})(2 - \sqrt{x})}\,dx} = 2\int{\left[\frac{1}{(1-\sqrt{x})(2 - \sqrt{x})}\right]\frac{1}{2\sqrt{x}}\,dx}

    \displaystyle = 2\int{\frac{1}{(1 - u)(2 - u)}\,du}

    which can then be solved using partial fractions. Mathaddict is correct.
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  8. #8
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    So the \sqrt{x} that is part of dx doesn't form part of the new integral!?
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  9. #9
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    \displaystyle \frac{1}{2\sqrt{x}}\,dx is replaced with \displaystyle du.

    Alternatively, \displaystyle \frac{1}{2\sqrt{x}} is replaced with \displaystyle \frac{du}{dx}, so that \displaystyle \frac{1}{2\sqrt{x}}\,dx = \frac{du}{dx}\,dx = du.

    But yes, the derivative that you find does not become part of the resulting \displaystyle u integral.
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  10. #10
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    Ah that does make sense. Many thanks!
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