Integral of $$$e^{- 3 x}$$$

Let $$$u=- 3 x$$$.

Then $$$du=\left(- 3 x\right)^{\prime }dx = - 3 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{3}$$$.

Thus,

$${\color{red}{\int{e^{- 3 x} d x}}} = {\color{red}{\int{\left(- \frac{e^{u}}{3}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{3}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{e^{u}}{3}\right)d u}}} = {\color{red}{\left(- \frac{\int{e^{u} d u}}{3}\right)}}$$

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

$$- \frac{{\color{red}{\int{e^{u} d u}}}}{3} = - \frac{{\color{red}{e^{u}}}}{3}$$

Recall that $$$u=- 3 x$$$:

$$- \frac{e^{{\color{red}{u}}}}{3} = - \frac{e^{{\color{red}{\left(- 3 x\right)}}}}{3}$$

Therefore,

$$\int{e^{- 3 x} d x} = - \frac{e^{- 3 x}}{3}$$

Add the constant of integration:

$$\int{e^{- 3 x} d x} = - \frac{e^{- 3 x}}{3}+C$$

Answer: $$$\int{e^{- 3 x} d x}=- \frac{e^{- 3 x}}{3}+C$$$