What is the integral of 1/(3^{-x)} ?

The integral of 1/(3^{-x)} is (3^x)/(ln(3))+C

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integral of 1/(3^{-x)}

\int \frac{1}{3 ^{- x}} d x

\frac{3 ^{x}}{ln ( 3 )} + C

Solution steps

\int \frac{1}{3 ^{- x}} d x

Simplify

\frac{1}{3 ^{- x}} : 3^{x}

= \int 3^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{3 ^{x}}{ln ( 3 )}

Add a constant to the solution

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

\int cos (5 x + 7) d x

f (x) = \frac{x ^{2} - 2}{x ^{2} + 2}

\frac{\partial}{\partial x} ((x + y) ln (\frac{x}{y}))

\int_{0}^{1} \frac{47}{4 y - 1} d y

y^{^{'}} = 3 - 6 x + y - 2 x y

What is the integral of 1/(3^{-x)} ?
The integral of 1/(3^{-x)} is (3^x)/(ln(3))+C