Integral de $$$\frac{1}{8 x - 3}$$$

Halla $$$\int \frac{1}{8 x - 3}\, dx$$$.

Sea $$$u=8 x - 3$$$.

Entonces $$$du=\left(8 x - 3\right)^{\prime }dx = 8 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{8}$$$.

Por lo tanto,

$${\color{red}{\int{\frac{1}{8 x - 3} d x}}} = {\color{red}{\int{\frac{1}{8 u} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{8}$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:

$${\color{red}{\int{\frac{1}{8 u} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{8}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

Recordemos que $$$u=8 x - 3$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} = \frac{\ln{\left(\left|{{\color{red}{\left(8 x - 3\right)}}}\right| \right)}}{8}$$

Por lo tanto,

$$\int{\frac{1}{8 x - 3} d x} = \frac{\ln{\left(\left|{8 x - 3}\right| \right)}}{8}$$

Añade la constante de integración:

$$\int{\frac{1}{8 x - 3} d x} = \frac{\ln{\left(\left|{8 x - 3}\right| \right)}}{8}+C$$

$$$\int \frac{1}{8 x - 3}\, dx = \frac{\ln\left(\left|{8 x - 3}\right|\right)}{8} + C$$$A