Integraal van $$$\frac{\ln\left(\sqrt{10} \sqrt{x}\right)}{x}$$$

Bepaal $$$\int \frac{\ln\left(\sqrt{10} \sqrt{x}\right)}{x}\, dx$$$.

Zij $$$u=\ln{\left(\sqrt{10} \sqrt{x} \right)}$$$.

Dan $$$du=\left(\ln{\left(\sqrt{10} \sqrt{x} \right)}\right)^{\prime }dx = \frac{1}{2 x} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\frac{dx}{x} = 2 du$$$.

Dus,

$${\color{red}{\int{\frac{\ln{\left(\sqrt{10} \sqrt{x} \right)}}{x} d x}}} = {\color{red}{\int{2 u d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=2$$$ en $$$f{\left(u \right)} = u$$$:

$${\color{red}{\int{2 u d u}}} = {\color{red}{\left(2 \int{u d u}\right)}}$$

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

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

We herinneren eraan dat $$$u=\ln{\left(\sqrt{10} \sqrt{x} \right)}$$$:

$${\color{red}{u}}^{2} = {\color{red}{\ln{\left(\sqrt{10} \sqrt{x} \right)}}}^{2}$$

Dus,

$$\int{\frac{\ln{\left(\sqrt{10} \sqrt{x} \right)}}{x} d x} = \ln{\left(\sqrt{10} \sqrt{x} \right)}^{2}$$

Vereenvoudig:

$$\int{\frac{\ln{\left(\sqrt{10} \sqrt{x} \right)}}{x} d x} = \frac{\left(\ln{\left(x \right)} + \ln{\left(10 \right)}\right)^{2}}{4}$$

Voeg de integratieconstante toe:

$$\int{\frac{\ln{\left(\sqrt{10} \sqrt{x} \right)}}{x} d x} = \frac{\left(\ln{\left(x \right)} + \ln{\left(10 \right)}\right)^{2}}{4}+C$$

$$$\int \frac{\ln\left(\sqrt{10} \sqrt{x}\right)}{x}\, dx = \frac{\left(\ln\left(x\right) + \ln\left(10\right)\right)^{2}}{4} + C$$$A