in general, how do you integrate a^x, a is constant?

Answer 1

You can rewrite
a^x = exp(ln(a^x)) = exp(ln(a)·x)

Therefore the integral is
∫ a^x dx = ∫ exp(ln(a)·x) dx = exp(ln(a)·x)/lna = a^x / ln(a)

Answer 2

Since e^(ln a) = a

∫ a^x dx = ∫ e^[(ln a)x] dx

u = (ln a) x
du = (ln a) dx

Substitute:

∫e^u [du/(ln a)] =

1/(ln a) ∫ e^u du

Solve:

1/(ln a) [e^u + C]

Substitute u = (ln a) x back into the equation:

1/(ln a) {e^[(ln a)x]} + C

1/(ln a) a^x + C =

Final answer:

(a^x)/(ln a) + C

Answer 3

we know d/dx(a^x) = a^x( in a)

integrate both sides a^x = int (a^x)( in a)

or int a^x = a^x/(In a) + C