Integral of $$$e^{9 x}$$$

Let $$$u=9 x$$$.

Then $$$du=\left(9 x\right)^{\prime }dx = 9 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{9}$$$.

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{9}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

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

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

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

Recall that $$$u=9 x$$$:

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

Therefore,

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

Add the constant of integration:

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

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