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September 22, 2014

September 22, 2014

Posted by **Ashley** on Friday, February 22, 2013 at 3:29pm.

x^2 divided by the square root of (2x-x^2) dx

- Calculus -
**Reiny**, Friday, February 22, 2013 at 4:24pmx^2/(2x - x^2)

= x/2 - 1

so int(x^2/(2x - x^2) dx

= int (x/2 - 1 ) dx

= (1/4)x^2 - x + c

- Calculus -
**Steve**, Friday, February 22, 2013 at 4:58pmeh? That's some fancy division there.

I get

2/(2-x) - 1

so the integral is

-2 ln(2-x) - x

or some equivalent

- Calculus -
**Steve**, Friday, February 22, 2013 at 5:14pmOops. I missed that pesky square root.

Let u = x-1 and we have

integral (u+1)^2/√(1-u^2) du

Now if u = sinθ

du = cosθ dθ

and we have integral of

(1+sinθ)^2 dθ

3/2 θ - 1/4 sin2θ - 2cosθ

= 3/2 arcsin(u) - 1/2 u√(1-u^2) - 2√(1-u^2)

= 3/2 arcsin(x-1) - 1/2 (x-1)√(2x-x^2) - 2√(2x-x^2)

= 1/2 (3arcsin(x-1) - (x+3)√(2x-x^2))

- Calculus -
**Reiny**, Friday, February 22, 2013 at 5:30pmOhhh my!!!

Mea Culpa!!!

That is embarrassing! What was I thinking??

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