Integraal van $$$\frac{\left(\sqrt{x} - 1\right)^{2}}{\sqrt{x}}$$$

Bepaal $$$\int \frac{\left(\sqrt{x} - 1\right)^{2}}{\sqrt{x}}\, dx$$$.

Zij $$$u=\sqrt{x} - 1$$$.

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

De integraal kan worden herschreven als

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

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

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

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

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

We herinneren eraan dat $$$u=\sqrt{x} - 1$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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