Integral of $$$9 e^{\frac{x}{4}}$$$

Find $$$\int 9 e^{\frac{x}{4}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=9$$$ and $$$f{\left(x \right)} = e^{\frac{x}{4}}$$$:

$${\color{red}{\int{9 e^{\frac{x}{4}} d x}}} = {\color{red}{\left(9 \int{e^{\frac{x}{4}} d x}\right)}}$$

Let $$$u=\frac{x}{4}$$$.

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

Therefore,

$$9 {\color{red}{\int{e^{\frac{x}{4}} d x}}} = 9 {\color{red}{\int{4 e^{u} d u}}}$$

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

$$9 {\color{red}{\int{4 e^{u} d u}}} = 9 {\color{red}{\left(4 \int{e^{u} d u}\right)}}$$

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

$$36 {\color{red}{\int{e^{u} d u}}} = 36 {\color{red}{e^{u}}}$$

Recall that $$$u=\frac{x}{4}$$$:

$$36 e^{{\color{red}{u}}} = 36 e^{{\color{red}{\left(\frac{x}{4}\right)}}}$$

Therefore,

$$\int{9 e^{\frac{x}{4}} d x} = 36 e^{\frac{x}{4}}$$

Add the constant of integration:

$$\int{9 e^{\frac{x}{4}} d x} = 36 e^{\frac{x}{4}}+C$$

$$$\int 9 e^{\frac{x}{4}}\, dx = 36 e^{\frac{x}{4}} + C$$$A