Calculus
posted by Ashley .
Find the integral of
x^2 divided by the square root of (2xx^2) dx

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.
I get
2/(2x)  1
so the integral is
2 ln(2x)  x
or some equivalent 
Oops. I missed that pesky square root.
Let u = x1 and we have
integral (u+1)^2/√(1u^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√(1u^2)  2√(1u^2)
= 3/2 arcsin(x1)  1/2 (x1)√(2xx^2)  2√(2xx^2)
= 1/2 (3arcsin(x1)  (x+3)√(2xx^2)) 
Ohhh my!!!
Mea Culpa!!!
That is embarrassing! What was I thinking??