Integral de $$$- \ln\left(x\right) + \frac{1}{\ln\left(x\right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \left(- \ln\left(x\right) + \frac{1}{\ln\left(x\right)}\right)\, dx$$$.
Solución
Integra término a término:
$${\color{red}{\int{\left(- \ln{\left(x \right)} + \frac{1}{\ln{\left(x \right)}}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{\ln{\left(x \right)}} d x} - \int{\ln{\left(x \right)} d x}\right)}}$$
Esta integral (Integral logarítmica) no tiene una forma cerrada:
$$- \int{\ln{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{\ln{\left(x \right)}} d x}}} = - \int{\ln{\left(x \right)} d x} + {\color{red}{\operatorname{li}{\left(x \right)}}}$$
Para la integral $$$\int{\ln{\left(x \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Sean $$$\operatorname{u}=\ln{\left(x \right)}$$$ y $$$\operatorname{dv}=dx$$$.
Entonces $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{1 d x}=x$$$ (los pasos pueden verse »).
Por lo tanto,
$$\operatorname{li}{\left(x \right)} - {\color{red}{\int{\ln{\left(x \right)} d x}}}=\operatorname{li}{\left(x \right)} - {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=\operatorname{li}{\left(x \right)} - {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$
Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:
$$- x \ln{\left(x \right)} + \operatorname{li}{\left(x \right)} + {\color{red}{\int{1 d x}}} = - x \ln{\left(x \right)} + \operatorname{li}{\left(x \right)} + {\color{red}{x}}$$
Por lo tanto,
$$\int{\left(- \ln{\left(x \right)} + \frac{1}{\ln{\left(x \right)}}\right)d x} = - x \ln{\left(x \right)} + x + \operatorname{li}{\left(x \right)}$$
Añade la constante de integración:
$$\int{\left(- \ln{\left(x \right)} + \frac{1}{\ln{\left(x \right)}}\right)d x} = - x \ln{\left(x \right)} + x + \operatorname{li}{\left(x \right)}+C$$
Respuesta
$$$\int \left(- \ln\left(x\right) + \frac{1}{\ln\left(x\right)}\right)\, dx = \left(- x \ln\left(x\right) + x + \operatorname{li}{\left(x \right)}\right) + C$$$A