Integral von $$$\frac{\ln^{4}\left(x\right)}{x}$$$

Bestimme $$$\int \frac{\ln^{4}\left(x\right)}{x}\, dx$$$.

Sei $$$u=\ln{\left(x \right)}$$$.

Dann $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (die Schritte sind » zu sehen), und es gilt $$$\frac{dx}{x} = du$$$.

Daher,

$${\color{red}{\int{\frac{\ln{\left(x \right)}^{4}}{x} d x}}} = {\color{red}{\int{u^{4} d u}}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=4$$$ an:

$${\color{red}{\int{u^{4} d u}}}={\color{red}{\frac{u^{1 + 4}}{1 + 4}}}={\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

Zur Erinnerung: $$$u=\ln{\left(x \right)}$$$:

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

Daher,

$$\int{\frac{\ln{\left(x \right)}^{4}}{x} d x} = \frac{\ln{\left(x \right)}^{5}}{5}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\ln{\left(x \right)}^{4}}{x} d x} = \frac{\ln{\left(x \right)}^{5}}{5}+C$$

$$$\int \frac{\ln^{4}\left(x\right)}{x}\, dx = \frac{\ln^{5}\left(x\right)}{5} + C$$$A