Integraal van $$$\frac{1}{- \sqrt{3} x + \sqrt{2} x}$$$

Bepaal $$$\int \frac{1}{- \sqrt{3} x + \sqrt{2} x}\, dx$$$.

Zij $$$u=- \sqrt{3} x + \sqrt{2} x$$$.

Dan $$$du=\left(- \sqrt{3} x + \sqrt{2} x\right)^{\prime }dx = \left(- \sqrt{3} + \sqrt{2}\right) dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{- \sqrt{3} + \sqrt{2}}$$$.

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{- \sqrt{3} + \sqrt{2}}$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

De integraal van $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

We herinneren eraan dat $$$u=- \sqrt{3} x + \sqrt{2} x$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{- \sqrt{3} + \sqrt{2}} = \frac{\ln{\left(\left|{{\color{red}{\left(- \sqrt{3} x + \sqrt{2} x\right)}}}\right| \right)}}{- \sqrt{3} + \sqrt{2}}$$

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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