# What is the integral of e^(3x)?

May 19, 2015

The answer is $\int {e}^{3 x} \mathrm{dx} = {e}^{3 x} / 3$.

So we have $f \left(x\right) = {e}^{3 x} = g \left(h \left(x\right)\right)$, where $g \left(x\right) = {e}^{x}$ and $h \left(x\right) = 3 x$.

The antiderivative of such a form is given by :

$\int g \left(h \left(x\right)\right) \cdot h ' \left(x\right) \mathrm{dx} = G \left(h \left(x\right)\right)$

We know that the derivative of $h \left(x\right) = 3 x$ is $h ' \left(x\right) = 3$.

We also know that the antiderivative of $g \left(x\right) = {e}^{x}$ is $G \left(x\right) = {e}^{x}$.

We have $\int {e}^{3 x} \mathrm{dx}$ but, with our formula, we can only calculate $\int {e}^{3 x} \cdot 3 \mathrm{dx}$, so what we will do is :

$\int {e}^{3 x} \mathrm{dx} = \frac{1}{3} \int {e}^{3 x} \cdot 3 \mathrm{dx} = {e}^{3 x} / 3$.

That's it.