Integral of $$$e^{4 u}$$$

Find $$$\int e^{4 u}\, du$$$.

Let $$$v=4 u$$$.

Then $$$dv=\left(4 u\right)^{\prime }du = 4 du$$$ (steps can be seen »), and we have that $$$du = \frac{dv}{4}$$$.

So,

$${\color{red}{\int{e^{4 u} d u}}} = {\color{red}{\int{\frac{e^{v}}{4} d v}}}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(v \right)} = e^{v}$$$:

$${\color{red}{\int{\frac{e^{v}}{4} d v}}} = {\color{red}{\left(\frac{\int{e^{v} d v}}{4}\right)}}$$

The integral of the exponential function is $$$\int{e^{v} d v} = e^{v}$$$:

$$\frac{{\color{red}{\int{e^{v} d v}}}}{4} = \frac{{\color{red}{e^{v}}}}{4}$$

Recall that $$$v=4 u$$$:

$$\frac{e^{{\color{red}{v}}}}{4} = \frac{e^{{\color{red}{\left(4 u\right)}}}}{4}$$

Therefore,

$$\int{e^{4 u} d u} = \frac{e^{4 u}}{4}$$

Add the constant of integration:

$$\int{e^{4 u} d u} = \frac{e^{4 u}}{4}+C$$

$$$\int e^{4 u}\, du = \frac{e^{4 u}}{4} + C$$$A