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

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

Partialbruchzerlegung durchführen:

$${\color{red}{\int{\frac{1}{a^{2} - x^{2}} d x}}} = {\color{red}{\int{\left(\frac{1}{2 a \left(a + x\right)} - \frac{1}{2 a \left(- a + x\right)}\right)d x}}}$$

Gliedweise integrieren:

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

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

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

Sei $$$u=a + x$$$.

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

Daher,

$$- \int{\frac{1}{2 a \left(- a + x\right)} d x} + \frac{{\color{red}{\int{\frac{1}{a + x} d x}}}}{2 a} = - \int{\frac{1}{2 a \left(- a + x\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2 a}$$

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

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

Zur Erinnerung: $$$u=a + x$$$:

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

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

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

Sei $$$u=- a + x$$$.

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

Das Integral lässt sich umschreiben als

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

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

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

Zur Erinnerung: $$$u=- a + x$$$:

$$\frac{\ln{\left(\left|{a + x}\right| \right)}}{2 a} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2 a} = \frac{\ln{\left(\left|{a + x}\right| \right)}}{2 a} - \frac{\ln{\left(\left|{{\color{red}{\left(- a + x\right)}}}\right| \right)}}{2 a}$$

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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