How do I integrate 3^x?

First I would ask what have they tried, and see if any of their attempts are on the right track - and then guide them from there, if not: 3^x is quite a difficult form to deal with, so is there any other way to represent this expression? wait for a time for the student to think/ respond What about if we made use of the exponential 'e'? (and its log 'ln')? wait for a time for the student to think/respond We know the exponential 'e' integrates very nicely: e^x goes to e^x (+c). We also know that e^(ln(f(x))) = f(x) because ln(x) is e(x)'s inverse operations - in the same way ((f(x) - 1) + 1) = f(x) because -1 and +1 are inverse operations! So how would we represent 3^x? wait for a time for the student to think/ respond Good job! If you let f(x) = 3^x, 3^x = e^(ln(3^x)) = e^(xln3) Where do we go from here? wait for a time for the student to think/ respond Excellent! Integration of  3^x = integration of e^(xln3), this is now a much more simple integration! wait for a time for the student to think/ respond Integration of e^(xln3) = (1/(ln3))*e^(xln3) (+c). Now do you think we can make this expression look a little more like the original? wait for a time for the student to think/ respond Yes, well done, since 3^x = e^(xln3), 1/(ln3))*e^(xln3) (+c) = 1/(ln3))*3^x (+c)! There you have it: integration of 3^x = 1/(ln3))*3^x (+c). Nice and neat! Can you generalise this for all a^x (a being a constant)?