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

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

Sea $$$u=x - 1$$$.

Entonces $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

Por lo tanto,

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

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

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

Recordemos que $$$u=x - 1$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}$$

Por lo tanto,

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

Añade la constante de integración:

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

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