Integral of $$$- 7 e^{- 7 x}$$$

Find $$$\int \left(- 7 e^{- 7 x}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-7$$$ and $$$f{\left(x \right)} = e^{- 7 x}$$$:

$${\color{red}{\int{\left(- 7 e^{- 7 x}\right)d x}}} = {\color{red}{\left(- 7 \int{e^{- 7 x} d x}\right)}}$$

Let $$$u=- 7 x$$$.

Then $$$du=\left(- 7 x\right)^{\prime }dx = - 7 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{7}$$$.

The integral can be rewritten as

$$- 7 {\color{red}{\int{e^{- 7 x} d x}}} = - 7 {\color{red}{\int{\left(- \frac{e^{u}}{7}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{7}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$- 7 {\color{red}{\int{\left(- \frac{e^{u}}{7}\right)d u}}} = - 7 {\color{red}{\left(- \frac{\int{e^{u} d u}}{7}\right)}}$$

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

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

Recall that $$$u=- 7 x$$$:

$$e^{{\color{red}{u}}} = e^{{\color{red}{\left(- 7 x\right)}}}$$

Therefore,

$$\int{\left(- 7 e^{- 7 x}\right)d x} = e^{- 7 x}$$

Add the constant of integration:

$$\int{\left(- 7 e^{- 7 x}\right)d x} = e^{- 7 x}+C$$

$$$\int \left(- 7 e^{- 7 x}\right)\, dx = e^{- 7 x} + C$$$A