Evaluate the Indefinite integral:

∫3 ((3x^2)+x-2)dx
0

(x^3 + x^2/2 -2x) from 0 to 3

(27 + 9/2 -6)

= (51/2) = 25.5

The constant can be taken out of the integral:
3 ∫ (3x^2) + x - 2 dx
Integrate, acknowledging that ∫ x^n dx = (x^(n+1)/(n+1)) + constant:
3 [(3x^3)/3 + (x^2)/2 - 2x + constant]
Simplify:
3[(x^3) + (x^2)/2 - 2x + constant]

Multiply out common factor if necessary. Hope this helps.

EDIT: HOWEVER if the 3 and the 0 in your question are limits, then may I remind you that integrals with limits are DEFINITE, not INDEFINITE!

∫ (3x^2) + x - 2 dx between limits 3 and 0
x^3 + (x^2)/2 - 2x between limits 3 and 0
[3^3 + (3^2)/2 - (2*3)] - [0^3 + (0^2)/2 - (2*0)]
27 + (9/2) - 6
= 25.5