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

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

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

integral

\frac{1}{e ^{- x}}

e^{x} + C

Solution steps

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

Simplify

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

= \int e^{x} d x

Use the common integral:

\int e^{x} d x = e^{x}

= e^{x}

Add a constant to the solution

= e^{x} + C

in t e g r a l \frac{1}{sec ( x )}

in t e g r a l \frac{1}{( x ^{2} - x )}

in t e g r a l \frac{( 2 x + 1 )}{( x ^{2} + 1 )}

in t e g r a l sin (2 x - y)

in t e g r a l \frac{1}{2 sin ( 2 x )}