Integralen av $$$\left(3 x + 1\right)^{3}$$$

Bestäm $$$\int \left(3 x + 1\right)^{3}\, dx$$$.

Låt $$$u=3 x + 1$$$ vara.

Då $$$du=\left(3 x + 1\right)^{\prime }dx = 3 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{3}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{3}$$$ och $$$f{\left(u \right)} = u^{3}$$$:

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=3$$$:

$$\frac{{\color{red}{\int{u^{3} d u}}}}{3}=\frac{{\color{red}{\frac{u^{1 + 3}}{1 + 3}}}}{3}=\frac{{\color{red}{\left(\frac{u^{4}}{4}\right)}}}{3}$$

Kom ihåg att $$$u=3 x + 1$$$:

$$\frac{{\color{red}{u}}^{4}}{12} = \frac{{\color{red}{\left(3 x + 1\right)}}^{4}}{12}$$

Alltså,

$$\int{\left(3 x + 1\right)^{3} d x} = \frac{\left(3 x + 1\right)^{4}}{12}$$

Lägg till integrationskonstanten:

$$\int{\left(3 x + 1\right)^{3} d x} = \frac{\left(3 x + 1\right)^{4}}{12}+C$$

$$$\int \left(3 x + 1\right)^{3}\, dx = \frac{\left(3 x + 1\right)^{4}}{12} + C$$$A