How do you evaluate the integral $\int 3^{x} d x$ $int3^(x) dx$ ?

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Answer 1 · 2018-03-14T22:26:26

$\int 3^{x} d x = \frac{1}{ln 3} 3^{x} + c$ $int3^xdx=1/ln3 3^x+"c"$

We want to find $\int 3^{x} d x$ $int3^xdx$ .

Make the natural substitution $u = 3^{x}$ $u=3^x$ so $d u = 3^{x} ln 3 d x$ $du=3^xln3dx$ .

So

$\int 3^{x} d x = \frac{1}{ln 3} \int 1 d u = \frac{1}{ln 3} u + c = \frac{1}{ln 3} 3^{x} + c$ $int3^xdx=1/ln3int1du=1/ln3 u+"c"=1/ln3 3^x+"c"$

How do you evaluate the integral ∫3xdxint3^(x) dx?