Integral von $$$\frac{e^{x}}{16 - 9 e^{2 x}}$$$

Bestimme $$$\int \frac{e^{x}}{16 - 9 e^{2 x}}\, dx$$$.

Sei $$$u=e^{x}$$$.

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

Das Integral lässt sich umschreiben als

$${\color{red}{\int{\frac{e^{x}}{16 - 9 e^{2 x}} d x}}} = {\color{red}{\int{\left(- \frac{1}{9 u^{2} - 16}\right)d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = \frac{1}{9 u^{2} - 16}$$$ an:

$${\color{red}{\int{\left(- \frac{1}{9 u^{2} - 16}\right)d u}}} = {\color{red}{\left(- \int{\frac{1}{9 u^{2} - 16} d u}\right)}}$$

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

$$- {\color{red}{\int{\frac{1}{9 u^{2} - 16} d u}}} = - {\color{red}{\int{\left(- \frac{1}{8 \left(3 u + 4\right)} + \frac{1}{8 \left(3 u - 4\right)}\right)d u}}}$$

Gliedweise integrieren:

$$- {\color{red}{\int{\left(- \frac{1}{8 \left(3 u + 4\right)} + \frac{1}{8 \left(3 u - 4\right)}\right)d u}}} = - {\color{red}{\left(\int{\frac{1}{8 \left(3 u - 4\right)} d u} - \int{\frac{1}{8 \left(3 u + 4\right)} d u}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{8}$$$ und $$$f{\left(u \right)} = \frac{1}{3 u + 4}$$$ an:

$$- \int{\frac{1}{8 \left(3 u - 4\right)} d u} + {\color{red}{\int{\frac{1}{8 \left(3 u + 4\right)} d u}}} = - \int{\frac{1}{8 \left(3 u - 4\right)} d u} + {\color{red}{\left(\frac{\int{\frac{1}{3 u + 4} d u}}{8}\right)}}$$

Sei $$$v=3 u + 4$$$.

Dann $$$dv=\left(3 u + 4\right)^{\prime }du = 3 du$$$ (die Schritte sind » zu sehen), und es gilt $$$du = \frac{dv}{3}$$$.

Also,

$$- \int{\frac{1}{8 \left(3 u - 4\right)} d u} + \frac{{\color{red}{\int{\frac{1}{3 u + 4} d u}}}}{8} = - \int{\frac{1}{8 \left(3 u - 4\right)} d u} + \frac{{\color{red}{\int{\frac{1}{3 v} d v}}}}{8}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(v \right)} = \frac{1}{v}$$$ an:

$$- \int{\frac{1}{8 \left(3 u - 4\right)} d u} + \frac{{\color{red}{\int{\frac{1}{3 v} d v}}}}{8} = - \int{\frac{1}{8 \left(3 u - 4\right)} d u} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{3}\right)}}}{8}$$

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

$$- \int{\frac{1}{8 \left(3 u - 4\right)} d u} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{24} = - \int{\frac{1}{8 \left(3 u - 4\right)} d u} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{24}$$

Zur Erinnerung: $$$v=3 u + 4$$$:

$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{24} - \int{\frac{1}{8 \left(3 u - 4\right)} d u} = \frac{\ln{\left(\left|{{\color{red}{\left(3 u + 4\right)}}}\right| \right)}}{24} - \int{\frac{1}{8 \left(3 u - 4\right)} d u}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{8}$$$ und $$$f{\left(u \right)} = \frac{1}{3 u - 4}$$$ an:

$$\frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - {\color{red}{\int{\frac{1}{8 \left(3 u - 4\right)} d u}}} = \frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - {\color{red}{\left(\frac{\int{\frac{1}{3 u - 4} d u}}{8}\right)}}$$

Sei $$$v=3 u - 4$$$.

Dann $$$dv=\left(3 u - 4\right)^{\prime }du = 3 du$$$ (die Schritte sind » zu sehen), und es gilt $$$du = \frac{dv}{3}$$$.

Daher,

$$\frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - \frac{{\color{red}{\int{\frac{1}{3 u - 4} d u}}}}{8} = \frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - \frac{{\color{red}{\int{\frac{1}{3 v} d v}}}}{8}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(v \right)} = \frac{1}{v}$$$ an:

$$\frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - \frac{{\color{red}{\int{\frac{1}{3 v} d v}}}}{8} = \frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{v} d v}}{3}\right)}}}{8}$$

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

$$\frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{24} = \frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{24}$$

Zur Erinnerung: $$$v=3 u - 4$$$:

$$\frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{24} = \frac{\ln{\left(\left|{3 u + 4}\right| \right)}}{24} - \frac{\ln{\left(\left|{{\color{red}{\left(3 u - 4\right)}}}\right| \right)}}{24}$$

Zur Erinnerung: $$$u=e^{x}$$$:

$$- \frac{\ln{\left(\left|{-4 + 3 {\color{red}{u}}}\right| \right)}}{24} + \frac{\ln{\left(\left|{4 + 3 {\color{red}{u}}}\right| \right)}}{24} = - \frac{\ln{\left(\left|{-4 + 3 {\color{red}{e^{x}}}}\right| \right)}}{24} + \frac{\ln{\left(\left|{4 + 3 {\color{red}{e^{x}}}}\right| \right)}}{24}$$

Daher,

$$\int{\frac{e^{x}}{16 - 9 e^{2 x}} d x} = \frac{\ln{\left(3 e^{x} + 4 \right)}}{24} - \frac{\ln{\left(\left|{3 e^{x} - 4}\right| \right)}}{24}$$

Vereinfachen:

$$\int{\frac{e^{x}}{16 - 9 e^{2 x}} d x} = \frac{\ln{\left(3 e^{x} + 4 \right)} - \ln{\left(\left|{3 e^{x} - 4}\right| \right)}}{24}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{e^{x}}{16 - 9 e^{2 x}} d x} = \frac{\ln{\left(3 e^{x} + 4 \right)} - \ln{\left(\left|{3 e^{x} - 4}\right| \right)}}{24}+C$$

$$$\int \frac{e^{x}}{16 - 9 e^{2 x}}\, dx = \frac{\ln\left(3 e^{x} + 4\right) - \ln\left(\left|{3 e^{x} - 4}\right|\right)}{24} + C$$$A