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

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

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

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

- e^{1 - x} + C

Solution steps

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

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= e (- \frac{1}{u})

Substitute back

u = e^{x}

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

Simplify

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

= - e^{1 - x}

Add a constant to the solution

= - e^{1 - x} + C

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