Integralen av $$$x \left(x^{2} - 1\right)$$$

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

Låt $$$u=x^{2} - 1$$$ vara.

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

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = u$$$:

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

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

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

Kom ihåg att $$$u=x^{2} - 1$$$:

$$\frac{{\color{red}{u}}^{2}}{4} = \frac{{\color{red}{\left(x^{2} - 1\right)}}^{2}}{4}$$

Alltså,

$$\int{x \left(x^{2} - 1\right) d x} = \frac{\left(x^{2} - 1\right)^{2}}{4}$$

Lägg till integrationskonstanten:

$$\int{x \left(x^{2} - 1\right) d x} = \frac{\left(x^{2} - 1\right)^{2}}{4}+C$$

$$$\int x \left(x^{2} - 1\right)\, dx = \frac{\left(x^{2} - 1\right)^{2}}{4} + C$$$A