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 |