求解： ∫(x2+2x-1/2x3+3x2-2x)dx

∫(x^2+2x-1/2x^3+3x^2-2x)dx

= ∫(4x^2 - 1/2x^3) dx Simplify the equation.

= (4x^3)/3 - (1/2x^4)/4 + C Integrate.

= 4/3x^3 - 1/8x^4 + C Simplify.

When we calculate this kind of integration, we must remember some formulae.
∫x^n dx=x^(n+1)/(n+1) +C where n is not equal to -1 and C is a constant.
When n=-1,
∫x^n dx=∫dx/x=ln x+C
∫n f(x) dx=n ∫f(x) dx
∫[f(x)+g(x)+m(x)+n(x)+...+y(x)]dx=∫f(x)dx+∫g(x)dx+∫m(x)dx+∫n(x)dx+...+∫y(x)dx

∫(x^2+2x-1/2x^3+3x^2-2x)dx
=∫(x^2+2x-x^(-3)/2+3x^2-2x)dx
=x^3/3+2x^2/2-[x^(-2)/(-2)]/2+3x^3/3-2x^2/2+C
=x^3/3+x^2+x^(-2)/4+x^3+x^2+C
=x^3/3+x^2+1/4x^2+x^3+x^2+C

Notice: x^n means x to the power of x.