# What is the integral of e^(7x)?

Dec 19, 2014

It's $\frac{1}{7} {e}^{7 x}$
What you want to calculate is:
$\int {e}^{7 x} \mathrm{dx}$

We're going to use u-substitution .

Let $u = 7 x$
Differentiate (derivative) both parts:
$\mathrm{du} = 7 \mathrm{dx}$
$\frac{\mathrm{du}}{7} = \mathrm{dx}$
Now we can replace everything in the integral:
$\int \frac{1}{7} {e}^{u} \mathrm{du}$
Bring the constant upfront
$\frac{1}{7} \int {e}^{u} \mathrm{du}$
The integral of ${e}^{u}$ is simply ${e}^{u}$
$\frac{1}{7} {e}^{u}$
And replace the $u$ back
$\frac{1}{7} {e}^{7 x}$

There's also a shortcut you can use:
Whenever you have a function of which you know the integral $f \left(x\right)$, but it has a different argument
$\implies$ the function is in the form $f \left(a x + b\right)$
If you want to integrate this, it is always equal to $\frac{1}{a} \cdot F \left(a x + b\right)$, where $F$ is the integral of the regular $f \left(x\right)$ function.

In this case:
$f \left(x\right) = {e}^{x}$
$F \left(x\right) = \int {e}^{x} \mathrm{dx} = {e}^{x}$
$a = 7$
$b = 0$
$f \left(a x + b\right) = {e}^{7 x}$
=> $\int {e}^{7 x} \mathrm{dx} = \frac{1}{a} \cdot F \left(a x + b\right) = \frac{1}{7} \cdot {e}^{7 x}$

If you use it more often, you will be able to do all these steps in your head.
Good luck!