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

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

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integral of e^{(x-1)}

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

e^{x - 1} + C

Solution steps

\int e^{x - 1} d x

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

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Use the common integral:

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

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

Simplify

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

= e^{x - 1}

Add a constant to the solution

= e^{x - 1} + C

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