Integralen av $$$x \left(2 x - 5\right)^{8}$$$

Bestäm $$$\int x \left(2 x - 5\right)^{8}\, dx$$$.

Låt $$$u=2 x - 5$$$ vara.

Då $$$du=\left(2 x - 5\right)^{\prime }dx = 2 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{2}$$$.

Integralen blir

$${\color{red}{\int{x \left(2 x - 5\right)^{8} d x}}} = {\color{red}{\int{\frac{u^{8} \left(u + 5\right)}{4} d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{4}$$$ och $$$f{\left(u \right)} = u^{8} \left(u + 5\right)$$$:

$${\color{red}{\int{\frac{u^{8} \left(u + 5\right)}{4} d u}}} = {\color{red}{\left(\frac{\int{u^{8} \left(u + 5\right) d u}}{4}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{u^{8} \left(u + 5\right) d u}}}}{4} = \frac{{\color{red}{\int{\left(u^{9} + 5 u^{8}\right)d u}}}}{4}$$

Integrera termvis:

$$\frac{{\color{red}{\int{\left(u^{9} + 5 u^{8}\right)d u}}}}{4} = \frac{{\color{red}{\left(\int{5 u^{8} d u} + \int{u^{9} d u}\right)}}}{4}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=9$$$:

$$\frac{\int{5 u^{8} d u}}{4} + \frac{{\color{red}{\int{u^{9} d u}}}}{4}=\frac{\int{5 u^{8} d u}}{4} + \frac{{\color{red}{\frac{u^{1 + 9}}{1 + 9}}}}{4}=\frac{\int{5 u^{8} d u}}{4} + \frac{{\color{red}{\left(\frac{u^{10}}{10}\right)}}}{4}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=5$$$ och $$$f{\left(u \right)} = u^{8}$$$:

$$\frac{u^{10}}{40} + \frac{{\color{red}{\int{5 u^{8} d u}}}}{4} = \frac{u^{10}}{40} + \frac{{\color{red}{\left(5 \int{u^{8} d u}\right)}}}{4}$$

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=8$$$:

$$\frac{u^{10}}{40} + \frac{5 {\color{red}{\int{u^{8} d u}}}}{4}=\frac{u^{10}}{40} + \frac{5 {\color{red}{\frac{u^{1 + 8}}{1 + 8}}}}{4}=\frac{u^{10}}{40} + \frac{5 {\color{red}{\left(\frac{u^{9}}{9}\right)}}}{4}$$

Kom ihåg att $$$u=2 x - 5$$$:

$$\frac{5 {\color{red}{u}}^{9}}{36} + \frac{{\color{red}{u}}^{10}}{40} = \frac{5 {\color{red}{\left(2 x - 5\right)}}^{9}}{36} + \frac{{\color{red}{\left(2 x - 5\right)}}^{10}}{40}$$

Alltså,

$$\int{x \left(2 x - 5\right)^{8} d x} = \frac{\left(2 x - 5\right)^{10}}{40} + \frac{5 \left(2 x - 5\right)^{9}}{36}$$

Förenkla:

$$\int{x \left(2 x - 5\right)^{8} d x} = \frac{\left(2 x - 5\right)^{9} \left(18 x + 5\right)}{360}$$

Lägg till integrationskonstanten:

$$\int{x \left(2 x - 5\right)^{8} d x} = \frac{\left(2 x - 5\right)^{9} \left(18 x + 5\right)}{360}+C$$

$$$\int x \left(2 x - 5\right)^{8}\, dx = \frac{\left(2 x - 5\right)^{9} \left(18 x + 5\right)}{360} + C$$$A