Integralen av $$$\frac{1}{x \ln^{2}\left(x\right)}$$$

Bestäm $$$\int \frac{1}{x \ln^{2}\left(x\right)}\, dx$$$.

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 $$$\frac{dx}{x} = du$$$.

Integralen kan omskrivas som

$${\color{red}{\int{\frac{1}{x \ln{\left(x \right)}^{2}} d x}}} = {\color{red}{\int{\frac{1}{u^{2}} d u}}}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

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

$$- {\color{red}{u}}^{-1} = - {\color{red}{\ln{\left(x \right)}}}^{-1}$$

Alltså,

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

Lägg till integrationskonstanten:

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

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