What is the integral of e^{ex} ?

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

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integral of e^{ex}

\int e^{e x} d x

e^{e x - 1} + C

Solution steps

\int e^{e x} d x

Apply u-substitution

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

e^{u - 1} = e^{- 1} e^{u}

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Use the common integral:

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

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

Substitute back

u = e x

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

Simplify

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

= e^{e x - 1}

Add a constant to the solution

= e^{e x - 1} + C