Integral von $$$\frac{f_{1} \tan^{2}{\left(f \right)}}{g}$$$ nach $$$g$$$
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Ihre Eingabe
Bestimme $$$\int \frac{f_{1} \tan^{2}{\left(f \right)}}{g}\, dg$$$.
Lösung
Wende die Konstantenfaktorregel $$$\int c f{\left(g \right)}\, dg = c \int f{\left(g \right)}\, dg$$$ mit $$$c=f_{1} \tan^{2}{\left(f \right)}$$$ und $$$f{\left(g \right)} = \frac{1}{g}$$$ an:
$${\color{red}{\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g}}} = {\color{red}{f_{1} \tan^{2}{\left(f \right)} \int{\frac{1}{g} d g}}}$$
Das Integral von $$$\frac{1}{g}$$$ ist $$$\int{\frac{1}{g} d g} = \ln{\left(\left|{g}\right| \right)}$$$:
$$f_{1} \tan^{2}{\left(f \right)} {\color{red}{\int{\frac{1}{g} d g}}} = f_{1} \tan^{2}{\left(f \right)} {\color{red}{\ln{\left(\left|{g}\right| \right)}}}$$
Daher,
$$\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g} = f_{1} \ln{\left(\left|{g}\right| \right)} \tan^{2}{\left(f \right)}$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\frac{f_{1} \tan^{2}{\left(f \right)}}{g} d g} = f_{1} \ln{\left(\left|{g}\right| \right)} \tan^{2}{\left(f \right)}+C$$
Antwort
$$$\int \frac{f_{1} \tan^{2}{\left(f \right)}}{g}\, dg = f_{1} \ln\left(\left|{g}\right|\right) \tan^{2}{\left(f \right)} + C$$$A