How do you integrate 7e^(2x) ?

Answer 1

∫7e^(2x)dx =

u = 2x
du = 2dx
du/2 = dx

∫7e^(2x)dx = (7/2)∫e^u du = (7/2)e^u + c

∫7e^(2x)dx = (7/2)e^(2x) + c

₢

Answer 2

7e^(2x)
substitute
2x=k
2dx=dk
dx=dk/2
so it becomes
(7/2)*e^k dk
as e^x integral is e^x
the integral is (7/2)*e^k
substitute k=2x
so your final answer is (7/2)*e^2x

Answer 3

integral a.e^(px+q) dx = (a/p).e^(px+q) + c
where c is an arbitrary constant of integration.

Then for this case, int. 7.e^(2x) dx = (7/2).e^(2x) + c