How do you evaluate the integral $int3^(x) dx$ ?

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Answer 1 · 2018-03-14T22:26:26

$int3^xdx=1/ln3 3^x+"c"$

We want to find $int3^xdx$ .

Make the natural substitution $u=3^x$ so $du=3^xln3dx$ .

So

$int3^xdx=1/ln3int1du=1/ln3 u+"c"=1/ln3 3^x+"c"$

How do you evaluate the integral int3^(x) dxint3^(x) dx?