# Thread: Integral using Partial Fractions with Square Roots

1. ## 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 $\displaystyle \int\frac{1}{\sqrt{3x-x^2}},dx$

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.

2. Originally Posted by manufacturedba
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 $\displaystyle \int\frac{1}{\sqrt{3x-x^2}},dx$

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?

$\displaystyle \sqrt{3x-x^2} = \sqrt{\frac{9}{4} -( x-(3/2))^2}$

3. 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):

$\displaystyle \sqrt{3x-x^2} = \frac{A}{x} + \frac{Bx + C}{3+x}$

4. Oops, there wouldn't be a Bx + C, just B

5. Originally Posted by manufacturedba
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):

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

if you want to use partial fractions here:

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

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

Now use the method of partial fraction to decompose the above term.

6. partial fractions doesn't apply, you need to perform a trig. substitution or some other way to integrate.

7. The 1 became $\displaystyle \sqrt{3x+1}$ 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?

8. $\displaystyle \int{\frac{1}{\sqrt{3x-x^2}}}\:dx=\int{\frac{1}{\sqrt{9/4-{(x-3/2)}^2}}}\:dx=$

$\displaystyle (t=x-3/2)$

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

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