Integraal van $$$\frac{\left(2 x - 1\right)^{4}}{16105}$$$

Bepaal $$$\int \frac{\left(2 x - 1\right)^{4}}{16105}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{16105}$$$ en $$$f{\left(x \right)} = \left(2 x - 1\right)^{4}$$$:

$${\color{red}{\int{\frac{\left(2 x - 1\right)^{4}}{16105} d x}}} = {\color{red}{\left(\frac{\int{\left(2 x - 1\right)^{4} d x}}{16105}\right)}}$$

Zij $$$u=2 x - 1$$$.

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

Dus,

$$\frac{{\color{red}{\int{\left(2 x - 1\right)^{4} d x}}}}{16105} = \frac{{\color{red}{\int{\frac{u^{4}}{2} d u}}}}{16105}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(u \right)} = u^{4}$$$:

$$\frac{{\color{red}{\int{\frac{u^{4}}{2} d u}}}}{16105} = \frac{{\color{red}{\left(\frac{\int{u^{4} d u}}{2}\right)}}}{16105}$$

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=4$$$:

$$\frac{{\color{red}{\int{u^{4} d u}}}}{32210}=\frac{{\color{red}{\frac{u^{1 + 4}}{1 + 4}}}}{32210}=\frac{{\color{red}{\left(\frac{u^{5}}{5}\right)}}}{32210}$$

We herinneren eraan dat $$$u=2 x - 1$$$:

$$\frac{{\color{red}{u}}^{5}}{161050} = \frac{{\color{red}{\left(2 x - 1\right)}}^{5}}{161050}$$

Dus,

$$\int{\frac{\left(2 x - 1\right)^{4}}{16105} d x} = \frac{\left(2 x - 1\right)^{5}}{161050}$$

Voeg de integratieconstante toe:

$$\int{\frac{\left(2 x - 1\right)^{4}}{16105} d x} = \frac{\left(2 x - 1\right)^{5}}{161050}+C$$

$$$\int \frac{\left(2 x - 1\right)^{4}}{16105}\, dx = \frac{\left(2 x - 1\right)^{5}}{161050} + C$$$A