Integral von $$$\frac{8 x}{4 x^{2} - 5}$$$

Bestimme $$$\int \frac{8 x}{4 x^{2} - 5}\, dx$$$.

Sei $$$u=4 x^{2} - 5$$$.

Dann $$$du=\left(4 x^{2} - 5\right)^{\prime }dx = 8 x dx$$$ (die Schritte sind » zu sehen), und es gilt $$$x dx = \frac{du}{8}$$$.

Das Integral wird zu

$${\color{red}{\int{\frac{8 x}{4 x^{2} - 5} d x}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

Zur Erinnerung: $$$u=4 x^{2} - 5$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\left(4 x^{2} - 5\right)}}}\right| \right)}$$

Daher,

$$\int{\frac{8 x}{4 x^{2} - 5} d x} = \ln{\left(\left|{4 x^{2} - 5}\right| \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{8 x}{4 x^{2} - 5} d x} = \ln{\left(\left|{4 x^{2} - 5}\right| \right)}+C$$

$$$\int \frac{8 x}{4 x^{2} - 5}\, dx = \ln\left(\left|{4 x^{2} - 5}\right|\right) + C$$$A