Integral of $$$- 37 e^{x} + \frac{37}{x}$$$

Find $$$\int \left(- 37 e^{x} + \frac{37}{x}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(- 37 e^{x} + \frac{37}{x}\right)d x}}} = {\color{red}{\left(\int{\frac{37}{x} d x} - \int{37 e^{x} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=37$$$ and $$$f{\left(x \right)} = e^{x}$$$:

$$\int{\frac{37}{x} d x} - {\color{red}{\int{37 e^{x} d x}}} = \int{\frac{37}{x} d x} - {\color{red}{\left(37 \int{e^{x} d x}\right)}}$$

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

$$\int{\frac{37}{x} d x} - 37 {\color{red}{\int{e^{x} d x}}} = \int{\frac{37}{x} d x} - 37 {\color{red}{e^{x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=37$$$ and $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- 37 e^{x} + {\color{red}{\int{\frac{37}{x} d x}}} = - 37 e^{x} + {\color{red}{\left(37 \int{\frac{1}{x} d x}\right)}}$$

The integral of $$$\frac{1}{x}$$$ is $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

$$- 37 e^{x} + 37 {\color{red}{\int{\frac{1}{x} d x}}} = - 37 e^{x} + 37 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Therefore,

$$\int{\left(- 37 e^{x} + \frac{37}{x}\right)d x} = - 37 e^{x} + 37 \ln{\left(\left|{x}\right| \right)}$$

Add the constant of integration:

$$\int{\left(- 37 e^{x} + \frac{37}{x}\right)d x} = - 37 e^{x} + 37 \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \left(- 37 e^{x} + \frac{37}{x}\right)\, dx = \left(- 37 e^{x} + 37 \ln\left(\left|{x}\right|\right)\right) + C$$$A