Integral de $$$\frac{16 x - 9}{x}$$$

Halla $$$\int \frac{16 x - 9}{x}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\frac{16 x - 9}{x} d x}}} = {\color{red}{\int{\left(16 - \frac{9}{x}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(16 - \frac{9}{x}\right)d x}}} = {\color{red}{\left(\int{16 d x} - \int{\frac{9}{x} d x}\right)}}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=16$$$:

$$- \int{\frac{9}{x} d x} + {\color{red}{\int{16 d x}}} = - \int{\frac{9}{x} d x} + {\color{red}{\left(16 x\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=9$$$ y $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$16 x - {\color{red}{\int{\frac{9}{x} d x}}} = 16 x - {\color{red}{\left(9 \int{\frac{1}{x} d x}\right)}}$$

La integral de $$$\frac{1}{x}$$$ es $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

$$16 x - 9 {\color{red}{\int{\frac{1}{x} d x}}} = 16 x - 9 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Por lo tanto,

$$\int{\frac{16 x - 9}{x} d x} = 16 x - 9 \ln{\left(\left|{x}\right| \right)}$$

Añade la constante de integración:

$$\int{\frac{16 x - 9}{x} d x} = 16 x - 9 \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \frac{16 x - 9}{x}\, dx = \left(16 x - 9 \ln\left(\left|{x}\right|\right)\right) + C$$$A