Integralen av $$$\frac{1}{x \ln^{3}\left(x\right)}$$$

Bestäm $$$\int \frac{1}{x \ln^{3}\left(x\right)}\, dx$$$.

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

Då $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (stegen kan ses »), och vi har att $$$\frac{dx}{x} = du$$$.

Alltså,

$${\color{red}{\int{\frac{1}{x \ln{\left(x \right)}^{3}} d x}}} = {\color{red}{\int{\frac{1}{u^{3}} d u}}}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-3$$$:

$${\color{red}{\int{\frac{1}{u^{3}} d u}}}={\color{red}{\int{u^{-3} d u}}}={\color{red}{\frac{u^{-3 + 1}}{-3 + 1}}}={\color{red}{\left(- \frac{u^{-2}}{2}\right)}}={\color{red}{\left(- \frac{1}{2 u^{2}}\right)}}$$

Kom ihåg att $$$u=\ln{\left(x \right)}$$$:

$$- \frac{{\color{red}{u}}^{-2}}{2} = - \frac{{\color{red}{\ln{\left(x \right)}}}^{-2}}{2}$$

Alltså,

$$\int{\frac{1}{x \ln{\left(x \right)}^{3}} d x} = - \frac{1}{2 \ln{\left(x \right)}^{2}}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{x \ln{\left(x \right)}^{3}} d x} = - \frac{1}{2 \ln{\left(x \right)}^{2}}+C$$

$$$\int \frac{1}{x \ln^{3}\left(x\right)}\, dx = - \frac{1}{2 \ln^{2}\left(x\right)} + C$$$A