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

Bestimme $$$\int \frac{1}{a^{2} - u^{2}}\, du$$$.

Partialbruchzerlegung durchführen:

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

Gliedweise integrieren:

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

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

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

Sei $$$v=a + u$$$.

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

Also,

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

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

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

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

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

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

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

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

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

Das Integral wird zu

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

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

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

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

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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