Integral de $$$\frac{\ln^{3}\left(x\right)}{x^{2}}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \frac{\ln^{3}\left(x\right)}{x^{2}}\, dx$$$.
Solución
Sea $$$u=\frac{1}{x}$$$.
Entonces $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{dx}{x^{2}} = - du$$$.
La integral puede reescribirse como
$${\color{red}{\int{\frac{\ln{\left(x \right)}^{3}}{x^{2}} d x}}} = {\color{red}{\int{\ln{\left(u \right)}^{3} d u}}}$$
Para la integral $$$\int{\ln{\left(u \right)}^{3} d u}$$$, utiliza la integración por partes $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.
Sean $$$\operatorname{g}=\ln{\left(u \right)}^{3}$$$ y $$$\operatorname{dv}=du$$$.
Entonces $$$\operatorname{dg}=\left(\ln{\left(u \right)}^{3}\right)^{\prime }du=\frac{3 \ln{\left(u \right)}^{2}}{u} du$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{1 d u}=u$$$ (los pasos pueden verse »).
La integral puede reescribirse como
$${\color{red}{\int{\ln{\left(u \right)}^{3} d u}}}={\color{red}{\left(\ln{\left(u \right)}^{3} \cdot u-\int{u \cdot \frac{3 \ln{\left(u \right)}^{2}}{u} d u}\right)}}={\color{red}{\left(u \ln{\left(u \right)}^{3} - \int{3 \ln{\left(u \right)}^{2} d u}\right)}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=3$$$ y $$$f{\left(u \right)} = \ln{\left(u \right)}^{2}$$$:
$$u \ln{\left(u \right)}^{3} - {\color{red}{\int{3 \ln{\left(u \right)}^{2} d u}}} = u \ln{\left(u \right)}^{3} - {\color{red}{\left(3 \int{\ln{\left(u \right)}^{2} d u}\right)}}$$
Para la integral $$$\int{\ln{\left(u \right)}^{2} d u}$$$, utiliza la integración por partes $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.
Sean $$$\operatorname{g}=\ln{\left(u \right)}^{2}$$$ y $$$\operatorname{dv}=du$$$.
Entonces $$$\operatorname{dg}=\left(\ln{\left(u \right)}^{2}\right)^{\prime }du=\frac{2 \ln{\left(u \right)}}{u} du$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{1 d u}=u$$$ (los pasos pueden verse »).
La integral puede reescribirse como
$$u \ln{\left(u \right)}^{3} - 3 {\color{red}{\int{\ln{\left(u \right)}^{2} d u}}}=u \ln{\left(u \right)}^{3} - 3 {\color{red}{\left(\ln{\left(u \right)}^{2} \cdot u-\int{u \cdot \frac{2 \ln{\left(u \right)}}{u} d u}\right)}}=u \ln{\left(u \right)}^{3} - 3 {\color{red}{\left(u \ln{\left(u \right)}^{2} - \int{2 \ln{\left(u \right)} d u}\right)}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=2$$$ y $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:
$$u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 3 {\color{red}{\int{2 \ln{\left(u \right)} d u}}} = u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 3 {\color{red}{\left(2 \int{\ln{\left(u \right)} d u}\right)}}$$
Para la integral $$$\int{\ln{\left(u \right)} d u}$$$, utiliza la integración por partes $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.
Sean $$$\operatorname{g}=\ln{\left(u \right)}$$$ y $$$\operatorname{dv}=du$$$.
Entonces $$$\operatorname{dg}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{1 d u}=u$$$ (los pasos pueden verse »).
Entonces,
$$u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 {\color{red}{\int{\ln{\left(u \right)} d u}}}=u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 {\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}=u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 {\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}$$
Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:
$$u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 u \ln{\left(u \right)} - 6 {\color{red}{\int{1 d u}}} = u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 u \ln{\left(u \right)} - 6 {\color{red}{u}}$$
Recordemos que $$$u=\frac{1}{x}$$$:
$$- 6 {\color{red}{u}} + 6 {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} - 3 {\color{red}{u}} \ln{\left({\color{red}{u}} \right)}^{2} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)}^{3} = - 6 {\color{red}{\frac{1}{x}}} + 6 {\color{red}{\frac{1}{x}}} \ln{\left({\color{red}{\frac{1}{x}}} \right)} - 3 {\color{red}{\frac{1}{x}}} \ln{\left({\color{red}{\frac{1}{x}}} \right)}^{2} + {\color{red}{\frac{1}{x}}} \ln{\left({\color{red}{\frac{1}{x}}} \right)}^{3}$$
Por lo tanto,
$$\int{\frac{\ln{\left(x \right)}^{3}}{x^{2}} d x} = \frac{\ln{\left(\frac{1}{x} \right)}^{3}}{x} - \frac{3 \ln{\left(\frac{1}{x} \right)}^{2}}{x} + \frac{6 \ln{\left(\frac{1}{x} \right)}}{x} - \frac{6}{x}$$
Simplificar:
$$\int{\frac{\ln{\left(x \right)}^{3}}{x^{2}} d x} = \frac{- \ln{\left(x \right)}^{3} - 3 \ln{\left(x \right)}^{2} - 6 \ln{\left(x \right)} - 6}{x}$$
Añade la constante de integración:
$$\int{\frac{\ln{\left(x \right)}^{3}}{x^{2}} d x} = \frac{- \ln{\left(x \right)}^{3} - 3 \ln{\left(x \right)}^{2} - 6 \ln{\left(x \right)} - 6}{x}+C$$
Respuesta
$$$\int \frac{\ln^{3}\left(x\right)}{x^{2}}\, dx = \frac{- \ln^{3}\left(x\right) - 3 \ln^{2}\left(x\right) - 6 \ln\left(x\right) - 6}{x} + C$$$A