What is the integral of e^{t-1} ?

The integral of e^{t-1} is e^{t-1}+C

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integral of e^{t-1}

\int e^{t - 1} d t

e^{t - 1} + C

Solution steps

\int e^{t - 1} d t

e^{t - 1} = e^{- 1} e^{t}

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Use the common integral:

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

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

Simplify

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

= e^{t - 1}

Add a constant to the solution

= e^{t - 1} + C