Integral of $$$2 e^{x} - 10$$$

Find $$$\int \left(2 e^{x} - 10\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(2 e^{x} - 10\right)d x}}} = {\color{red}{\left(- \int{10 d x} + \int{2 e^{x} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=10$$$:

$$\int{2 e^{x} d x} - {\color{red}{\int{10 d x}}} = \int{2 e^{x} d x} - {\color{red}{\left(10 x\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = e^{x}$$$:

$$- 10 x + {\color{red}{\int{2 e^{x} d x}}} = - 10 x + {\color{red}{\left(2 \int{e^{x} d x}\right)}}$$

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

$$- 10 x + 2 {\color{red}{\int{e^{x} d x}}} = - 10 x + 2 {\color{red}{e^{x}}}$$

Therefore,

$$\int{\left(2 e^{x} - 10\right)d x} = - 10 x + 2 e^{x}$$

Add the constant of integration:

$$\int{\left(2 e^{x} - 10\right)d x} = - 10 x + 2 e^{x}+C$$

$$$\int \left(2 e^{x} - 10\right)\, dx = \left(- 10 x + 2 e^{x}\right) + C$$$A