Integral von $$$\frac{\left(3 x + 1\right)^{2}}{244}$$$

Bestimme $$$\int \frac{\left(3 x + 1\right)^{2}}{244}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{244}$$$ und $$$f{\left(x \right)} = \left(3 x + 1\right)^{2}$$$ an:

$${\color{red}{\int{\frac{\left(3 x + 1\right)^{2}}{244} d x}}} = {\color{red}{\left(\frac{\int{\left(3 x + 1\right)^{2} d x}}{244}\right)}}$$

Sei $$$u=3 x + 1$$$.

Dann $$$du=\left(3 x + 1\right)^{\prime }dx = 3 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{3}$$$.

Also,

$$\frac{{\color{red}{\int{\left(3 x + 1\right)^{2} d x}}}}{244} = \frac{{\color{red}{\int{\frac{u^{2}}{3} d u}}}}{244}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(u \right)} = u^{2}$$$ an:

$$\frac{{\color{red}{\int{\frac{u^{2}}{3} d u}}}}{244} = \frac{{\color{red}{\left(\frac{\int{u^{2} d u}}{3}\right)}}}{244}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=2$$$ an:

$$\frac{{\color{red}{\int{u^{2} d u}}}}{732}=\frac{{\color{red}{\frac{u^{1 + 2}}{1 + 2}}}}{732}=\frac{{\color{red}{\left(\frac{u^{3}}{3}\right)}}}{732}$$

Zur Erinnerung: $$$u=3 x + 1$$$:

$$\frac{{\color{red}{u}}^{3}}{2196} = \frac{{\color{red}{\left(3 x + 1\right)}}^{3}}{2196}$$

Daher,

$$\int{\frac{\left(3 x + 1\right)^{2}}{244} d x} = \frac{\left(3 x + 1\right)^{3}}{2196}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\left(3 x + 1\right)^{2}}{244} d x} = \frac{\left(3 x + 1\right)^{3}}{2196}+C$$

$$$\int \frac{\left(3 x + 1\right)^{2}}{244}\, dx = \frac{\left(3 x + 1\right)^{3}}{2196} + C$$$A