Integral von $$$\frac{4 x^{2} - 2 \sqrt{2} x}{x}$$$

Bestimme $$$\int \frac{4 x^{2} - 2 \sqrt{2} x}{x}\, dx$$$.

Expand the expression:

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

Gliedweise integrieren:

$${\color{red}{\int{\left(4 x - 2 \sqrt{2}\right)d x}}} = {\color{red}{\left(- \int{2 \sqrt{2} d x} + \int{4 x d x}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=2 \sqrt{2}$$$ an:

$$\int{4 x d x} - {\color{red}{\int{2 \sqrt{2} d x}}} = \int{4 x d x} - {\color{red}{\left(2 \sqrt{2} x\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=4$$$ und $$$f{\left(x \right)} = x$$$ an:

$$- 2 \sqrt{2} x + {\color{red}{\int{4 x d x}}} = - 2 \sqrt{2} x + {\color{red}{\left(4 \int{x d x}\right)}}$$

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

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

Daher,

$$\int{\frac{4 x^{2} - 2 \sqrt{2} x}{x} d x} = 2 x^{2} - 2 \sqrt{2} x$$

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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