Integral von $$$\frac{16 x - 9}{x}$$$

Bestimme $$$\int \frac{16 x - 9}{x}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\frac{16 x - 9}{x} d x}}} = {\color{red}{\int{\left(16 - \frac{9}{x}\right)d x}}}$$

Gliedweise integrieren:

$${\color{red}{\int{\left(16 - \frac{9}{x}\right)d x}}} = {\color{red}{\left(\int{16 d x} - \int{\frac{9}{x} d x}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=16$$$ an:

$$- \int{\frac{9}{x} d x} + {\color{red}{\int{16 d x}}} = - \int{\frac{9}{x} d x} + {\color{red}{\left(16 x\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=9$$$ und $$$f{\left(x \right)} = \frac{1}{x}$$$ an:

$$16 x - {\color{red}{\int{\frac{9}{x} d x}}} = 16 x - {\color{red}{\left(9 \int{\frac{1}{x} d x}\right)}}$$

Das Integral von $$$\frac{1}{x}$$$ ist $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

$$16 x - 9 {\color{red}{\int{\frac{1}{x} d x}}} = 16 x - 9 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Daher,

$$\int{\frac{16 x - 9}{x} d x} = 16 x - 9 \ln{\left(\left|{x}\right| \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{16 x - 9}{x} d x} = 16 x - 9 \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \frac{16 x - 9}{x}\, dx = \left(16 x - 9 \ln\left(\left|{x}\right|\right)\right) + C$$$A