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

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

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 blir

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

Integralen av $$$\frac{1}{u}$$$ är $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\ln{\left(x \right)}}}}\right| \right)}$$

Alltså,

$$\int{\frac{1}{x \ln{\left(x \right)}} d x} = \ln{\left(\left|{\ln{\left(x \right)}}\right| \right)}$$

Lägg till integrationskonstanten:

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

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