Integral von $$$\frac{x}{2 x^{2} - 1}$$$

Bestimme $$$\int \frac{x}{2 x^{2} - 1}\, dx$$$.

Sei $$$u=2 x^{2} - 1$$$.

Dann $$$du=\left(2 x^{2} - 1\right)^{\prime }dx = 4 x dx$$$ (die Schritte sind » zu sehen), und es gilt $$$x dx = \frac{du}{4}$$$.

Also,

$${\color{red}{\int{\frac{x}{2 x^{2} - 1} d x}}} = {\color{red}{\int{\frac{1}{4 u} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(u \right)} = \frac{1}{u}$$$ an:

$${\color{red}{\int{\frac{1}{4 u} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{4}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

Zur Erinnerung: $$$u=2 x^{2} - 1$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \frac{x}{2 x^{2} - 1}\, dx = \frac{\ln\left(\left|{2 x^{2} - 1}\right|\right)}{4} + C$$$A