How do you integrate #int 3^xdx#?

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Answer 1 · 2018-03-17T07:20:10

The answer is #=3^x/ln3+C#

Let #u=3^x#

Taking logs on both sides

#lnu=ln(3^x)=xln3#

Then

#u=e^(xln3)#

Therefore,

#int3^xdx=inte^(xln3)dx=e^(xln3)/(ln3)+C#

#=3^x/ln3+C#