Integraal van $$$\frac{1}{x \sqrt{\ln\left(x\right)}}$$$

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

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

Dan $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (de stappen zijn te zien »), en dan geldt dat $$$\frac{dx}{x} = du$$$.

De integraal wordt

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

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=- \frac{1}{2}$$$:

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

We herinneren eraan dat $$$u=\ln{\left(x \right)}$$$:

$$2 \sqrt{{\color{red}{u}}} = 2 \sqrt{{\color{red}{\ln{\left(x \right)}}}}$$

Dus,

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

Voeg de integratieconstante toe:

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

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