Let x²=4-y, then 2xdx=-dy and y=4-x². The integral becomes:
-∫(4-x²)x(2xdx)=-2∫(4x²-x⁴)dx=-2(4x³/3-x⁵/5)+c.
Now back-substitute for x:
-2x³(4/3-x²/5)+c=-2(4-y)√(4-y)(4/3-(4-y)/5)+c
-2(4-y)√(4-y)((20-3(4-y))/15)+c=-(2/15)(4-y)√(4-y)(8+3y)+c.
This can also be written: -(2/15)(8+3y)(4-y)^(3/2)+c, where c is a constant of integration.