Integral de $$$\ln\left(-1 + \frac{1}{x}\right)$$$

Halla $$$\int \ln\left(-1 + \frac{1}{x}\right)\, dx$$$.

Para la integral $$$\int{\ln{\left(-1 + \frac{1}{x} \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\ln{\left(-1 + \frac{1}{x} \right)}$$$ y $$$\operatorname{dv}=dx$$$.

Entonces $$$\operatorname{du}=\left(\ln{\left(-1 + \frac{1}{x} \right)}\right)^{\prime }dx=\frac{1}{x \left(x - 1\right)} dx$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{1 d x}=x$$$ (los pasos pueden verse »).

La integral se convierte en

$${\color{red}{\int{\ln{\left(-1 + \frac{1}{x} \right)} d x}}}={\color{red}{\left(\ln{\left(-1 + \frac{1}{x} \right)} \cdot x-\int{x \cdot \frac{1}{x \left(x - 1\right)} d x}\right)}}={\color{red}{\left(x \ln{\left(-1 + \frac{1}{x} \right)} - \int{\frac{1}{x - 1} d x}\right)}}$$

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

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

La integral se convierte en

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

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

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

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

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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