# Integrals of Exponential Functions

## Key Questions

• By rewriting a bit,

$\int {n}^{x} \mathrm{dx} = \int {e}^{\left(\ln n\right) x} \mathrm{dx} = {e}^{\left(\ln n\right) x} / \left\{\ln n\right\} + C = {n}^{x} / \left\{\ln n\right\} + C$

I hope that this was helpful.

• The answer is $I = - {e}^{- x} + C$.

This integral can be solved by a substitution:

$u = - x$
$\mathrm{du} = - \mathrm{dx}$
$- \mathrm{du} = \mathrm{dx}$

So, now we can substitute:

$\int {e}^{- x} \mathrm{dx} = \int {e}^{u} \left(- \mathrm{du}\right)$
$= - \int {e}^{u} \mathrm{du}$
$= - {e}^{u} + C$

and substitute back:
$= - {e}^{- x} + C$

For simple looking integrands, you should try a quick check to see if substitution works before trying harder integration methods.

$\int {3}^{x} \mathrm{dx} = \frac{1}{\ln} 3 {3}^{x} + \text{c}$

#### Explanation:

We want to find $\int {3}^{x} \mathrm{dx}$.

Make the natural substitution $u = {3}^{x}$ so $\mathrm{du} = {3}^{x} \ln 3 \mathrm{dx}$.

So

$\int {3}^{x} \mathrm{dx} = \frac{1}{\ln} 3 \int 1 \mathrm{du} = \frac{1}{\ln} 3 u + \text{c"=1/ln3 3^x+"c}$

• Since

$\left({e}^{x}\right) ' = {e}^{x}$,

we have

$\int {e}^{x} \mathrm{dx} = {e}^{x} + C$.

I hope that this was helpful.