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

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

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

integral

(e^{x})^{- 1}

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

Solution steps

\int (e^{x})^{- 1} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply u-substitution

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= - \frac{1}{u}

Substitute back

u = e^{x}

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

Add a constant to the solution

= - \frac{1}{e ^{x}} + C