Integralen av $$$\sqrt{x - 2} + 1$$$

Bestäm $$$\int \left(\sqrt{x - 2} + 1\right)\, dx$$$.

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=1$$$:

$$\int{\sqrt{x - 2} d x} + {\color{red}{\int{1 d x}}} = \int{\sqrt{x - 2} d x} + {\color{red}{x}}$$

Låt $$$u=x - 2$$$ vara.

Då $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Alltså,

$$x + {\color{red}{\int{\sqrt{x - 2} d x}}} = x + {\color{red}{\int{\sqrt{u} d u}}}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=\frac{1}{2}$$$:

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

Kom ihåg att $$$u=x - 2$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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