Integral von $$$\frac{4 \ln\left(2 x\right) - 9}{x}$$$

Bestimme $$$\int \frac{4 \ln\left(2 x\right) - 9}{x}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\frac{4 \ln{\left(2 x \right)} - 9}{x} d x}}} = {\color{red}{\int{\left(\frac{4 \ln{\left(x \right)}}{x} - \frac{9}{x} + \frac{4 \ln{\left(2 \right)}}{x}\right)d x}}}$$

Gliedweise integrieren:

$${\color{red}{\int{\left(\frac{4 \ln{\left(x \right)}}{x} - \frac{9}{x} + \frac{4 \ln{\left(2 \right)}}{x}\right)d x}}} = {\color{red}{\left(- \int{\frac{9}{x} d x} + \int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x}\right)}}$$

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:

$$\int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} - {\color{red}{\int{\frac{9}{x} d x}}} = \int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} - {\color{red}{\left(9 \int{\frac{1}{x} d x}\right)}}$$

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

$$\int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} - 9 {\color{red}{\int{\frac{1}{x} d x}}} = \int{\frac{4 \ln{\left(2 \right)}}{x} d x} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} - 9 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=4 \ln{\left(2 \right)}$$$ und $$$f{\left(x \right)} = \frac{1}{x}$$$ an:

$$- 9 \ln{\left(\left|{x}\right| \right)} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} + {\color{red}{\int{\frac{4 \ln{\left(2 \right)}}{x} d x}}} = - 9 \ln{\left(\left|{x}\right| \right)} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} + {\color{red}{\left(4 \ln{\left(2 \right)} \int{\frac{1}{x} d x}\right)}}$$

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

$$- 9 \ln{\left(\left|{x}\right| \right)} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} + 4 \ln{\left(2 \right)} {\color{red}{\int{\frac{1}{x} d x}}} = - 9 \ln{\left(\left|{x}\right| \right)} + \int{\frac{4 \ln{\left(x \right)}}{x} d x} + 4 \ln{\left(2 \right)} {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

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

$$- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + {\color{red}{\int{\frac{4 \ln{\left(x \right)}}{x} d x}}} = - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + {\color{red}{\left(4 \int{\frac{\ln{\left(x \right)}}{x} d x}\right)}}$$

Sei $$$u=\ln{\left(x \right)}$$$.

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

Das Integral lässt sich umschreiben als

$$- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\int{\frac{\ln{\left(x \right)}}{x} d x}}} = - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\int{u d u}}}$$

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

$$- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\int{u d u}}}=- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 4 {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

Zur Erinnerung: $$$u=\ln{\left(x \right)}$$$:

$$- 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 2 {\color{red}{u}}^{2} = - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)} + 2 {\color{red}{\ln{\left(x \right)}}}^{2}$$

Daher,

$$\int{\frac{4 \ln{\left(2 x \right)} - 9}{x} d x} = 2 \ln{\left(x \right)}^{2} - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{4 \ln{\left(2 x \right)} - 9}{x} d x} = 2 \ln{\left(x \right)}^{2} - 9 \ln{\left(\left|{x}\right| \right)} + 4 \ln{\left(2 \right)} \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \frac{4 \ln\left(2 x\right) - 9}{x}\, dx = \left(2 \ln^{2}\left(x\right) - 9 \ln\left(\left|{x}\right|\right) + 4 \ln\left(2\right) \ln\left(\left|{x}\right|\right)\right) + C$$$A