# Calculus

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

• Calculus -

x^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 -

eh? 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 -

Oops. 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 -

Ohhh my!!!

Mea Culpa!!!

That is embarrassing! What was I thinking??

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