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,

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

$=3^x/ln3+C$

How do you integrate int 3^xdxint 3^xdx?