Integral de $$$\frac{\ln^{5}\left(u^{2}\right)}{u}$$$

Halla $$$\int \frac{\ln^{5}\left(u^{2}\right)}{u}\, du$$$.

La entrada se reescribe: $$$\int{\frac{\ln{\left(u^{2} \right)}^{5}}{u} d u}=\int{\frac{32 \ln{\left(u \right)}^{5}}{u} d u}$$$.

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=32$$$ y $$$f{\left(u \right)} = \frac{\ln{\left(u \right)}^{5}}{u}$$$:

$${\color{red}{\int{\frac{32 \ln{\left(u \right)}^{5}}{u} d u}}} = {\color{red}{\left(32 \int{\frac{\ln{\left(u \right)}^{5}}{u} d u}\right)}}$$

Sea $$$v=\ln{\left(u \right)}$$$.

Entonces $$$dv=\left(\ln{\left(u \right)}\right)^{\prime }du = \frac{du}{u}$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{du}{u} = dv$$$.

Entonces,

$$32 {\color{red}{\int{\frac{\ln{\left(u \right)}^{5}}{u} d u}}} = 32 {\color{red}{\int{v^{5} d v}}}$$

Aplica la regla de la potencia $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=5$$$:

$$32 {\color{red}{\int{v^{5} d v}}}=32 {\color{red}{\frac{v^{1 + 5}}{1 + 5}}}=32 {\color{red}{\left(\frac{v^{6}}{6}\right)}}$$

Recordemos que $$$v=\ln{\left(u \right)}$$$:

$$\frac{16 {\color{red}{v}}^{6}}{3} = \frac{16 {\color{red}{\ln{\left(u \right)}}}^{6}}{3}$$

Por lo tanto,

$$\int{\frac{32 \ln{\left(u \right)}^{5}}{u} d u} = \frac{16 \ln{\left(u \right)}^{6}}{3}$$

Añade la constante de integración:

$$\int{\frac{32 \ln{\left(u \right)}^{5}}{u} d u} = \frac{16 \ln{\left(u \right)}^{6}}{3}+C$$

$$$\int \frac{\ln^{5}\left(u^{2}\right)}{u}\, du = \frac{16 \ln^{6}\left(u\right)}{3} + C$$$A