calculus IV

Sub u = xy, then u' = xy' + y
So the equation becomes:
(u/x)2 u' = x/√(1 + x4)
u2du = x3dx/√(1 + x4)
∫u2du =∫ x3dx/√(1 + x4)
u3/3 = (1/4)∫ d(1+ x4)/√(1 + x4)
u3 = (3/2)√(1 + x4) + C where C is a constant
x3y3 = (3/2)√(1 + x4) + C
y3 = 3√(1 + x4)/(2x3) + Cx-3
y = [3√(1 + x4)/(2x3) + Cx-3]1/3

2009-02-24 14:24:49 補充：
This is called the substitution method

This is just to put a variable in other form to make the equation to be solvable. This is a usual technique used in such calculus problem.
Then at last, indeed you have to put back the given variable to eliminate the variable you made.

where do u find "u = xy" ?