Integraal van $$$x \sqrt{x^{2} - 1}$$$

Bepaal $$$\int x \sqrt{x^{2} - 1}\, dx$$$.

Zij $$$u=x^{2} - 1$$$.

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

Dus,

$${\color{red}{\int{x \sqrt{x^{2} - 1} d x}}} = {\color{red}{\int{\frac{\sqrt{u}}{2} d u}}}$$

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

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

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

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

We herinneren eraan dat $$$u=x^{2} - 1$$$:

$$\frac{{\color{red}{u}}^{\frac{3}{2}}}{3} = \frac{{\color{red}{\left(x^{2} - 1\right)}}^{\frac{3}{2}}}{3}$$

Dus,

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

Voeg de integratieconstante toe:

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

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