There is a rule for integrating ekx dx, or you can use a simple substitution u = 2x. I'll go through the steps of the substitution here. Taking the derivative of u = 2x, we find du = 2dx. Substituting u for 2x and du/2 for dx, we get:

u = 2x;   du = 2dx
∫ 3e2x dx   =  3 ∫ eu (du/2)
=  (3/2) ∫ eu du
=  (3/2) eu  + C
=  (3/2) e2x + C