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Math Help - Integral using Partial Fractions with Square Roots

  1. #1
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    Integral using Partial Fractions with Square Roots

    New to the forums (Obviously)

    I'm having slight trouble with a problem relating to the volume of a rotating region R, but I only need help with the integral of the function itself.

    I have \int\frac{1}{\sqrt{3x-x^2}},dx

    I am required to use partial fractions and obviously substitution for the problem so no integral calculator answers please. I can work it out once I have the partial fractions used in solving.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by manufacturedba View Post
    New to the forums (Obviously)

    I'm having slight trouble with a problem relating to the volume of a rotating region R, but I only need help with the integral of the function itself.

    I have \int\frac{1}{\sqrt{3x-x^2}},dx

    I am required to use partial fractions and obviously substitution for the problem so no integral calculator answers please. I can work it out once I have the partial fractions used in solving.
    Partial fraction?

    how about completing the square?

    \sqrt{3x-x^2} = \sqrt{\frac{9}{4} -( x-(3/2))^2}
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  3. #3
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    Unfortunately, that does not simplify the problem. That leaves me with a more complicated denominator with subtracting components. I can't pull them out of the square root. The closest I have gotten to an answer (I think):

    \sqrt{3x-x^2} = \frac{A}{x} + \frac{Bx + C}{3+x}
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  4. #4
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    Oops, there wouldn't be a Bx + C, just B
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by manufacturedba View Post
    Unfortunately, that does not simplify the problem. That leaves me with a more complicated denominator with subtracting components. I can't pull them out of the square root. The closest I have gotten to an answer (I think):

    \sqrt{3x-x^2} = \frac{A}{x} + \frac{Bx + C}{3+x} Where did you 1 in the numerator go??


    if you want to use partial fractions here:

    then
     \frac{1}{\sqrt{3x-x^2}} = \frac{1}{\sqrt{3x-x^2}} \times \frac{\sqrt{3x-x^2}}{\sqrt{3x-x^2}}

    = \frac{\sqrt{3x-x^2}}{3x-x^2}

    Now use the method of partial fraction to decompose the above term.
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  6. #6
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    partial fractions doesn't apply, you need to perform a trig. substitution or some other way to integrate.
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  7. #7
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    The 1 became \sqrt{3x+1} just like yours did. I just didn't carry through multiplication of A and B and cancel out the denominator. You arrived at the same point as me.

    But what to do about the square root? Square the both sides?
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  8. #8
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    \int{\frac{1}{\sqrt{3x-x^2}}}\:dx=\int{\frac{1}{\sqrt{9/4-{(x-3/2)}^2}}}\:dx=

    (t=x-3/2)

    =\int{\frac{1}{\sqrt{{(3/2)}^2-t^2}}}\:dt= \arcsin{(2t/3)} \: +\:C
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