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
eh? That's some fancy division there.
2/(2-x) - 1
so the integral is
-2 ln(2-x) - x
or some equivalent
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
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))
That is embarrassing! What was I thinking??
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