How do you integrate $3^{x}$ $3^x$ ?

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Answer 1 · 2016-09-13T07:57:45

$= \frac{1}{ln 3} 3^{x} + C$ $= 1/(ln 3) 3^x + C$

We can work the derivative first

$y = 3^{x}$ $y = 3^x$

$ln y = x ln 3$ $ln y = x ln 3$

$1/y y' = ln 3$

$y' = ln 3 \ 3^x$

$implies int 3^x \ dx$

$= int d/dx(1/(ln 3) 3^x )\ dx$

$= 1/(ln 3) 3^x + C$

How do you integrate 3^x3x 3^x?