Integral von $$$- 8 x + \tan{\left(x \right)} \sec{\left(x \right)}$$$

Bestimme $$$\int \left(- 8 x + \tan{\left(x \right)} \sec{\left(x \right)}\right)\, dx$$$.

Gliedweise integrieren:

$${\color{red}{\int{\left(- 8 x + \tan{\left(x \right)} \sec{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{8 x d x} + \int{\tan{\left(x \right)} \sec{\left(x \right)} d x}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=8$$$ und $$$f{\left(x \right)} = x$$$ an:

$$\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} - {\color{red}{\int{8 x d x}}} = \int{\tan{\left(x \right)} \sec{\left(x \right)} d x} - {\color{red}{\left(8 \int{x d x}\right)}}$$

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

$$\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} - 8 {\color{red}{\int{x d x}}}=\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} - 8 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} - 8 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Das Integral von $$$\tan{\left(x \right)} \sec{\left(x \right)}$$$ ist $$$\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} = \sec{\left(x \right)}$$$:

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

Daher,

$$\int{\left(- 8 x + \tan{\left(x \right)} \sec{\left(x \right)}\right)d x} = - 4 x^{2} + \sec{\left(x \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(- 8 x + \tan{\left(x \right)} \sec{\left(x \right)}\right)d x} = - 4 x^{2} + \sec{\left(x \right)}+C$$

$$$\int \left(- 8 x + \tan{\left(x \right)} \sec{\left(x \right)}\right)\, dx = \left(- 4 x^{2} + \sec{\left(x \right)}\right) + C$$$A