Integral von $$$\frac{1}{- k^{2} + r^{2}}$$$ nach $$$r$$$

Bestimme $$$\int \frac{1}{- k^{2} + r^{2}}\, dr$$$.

Partialbruchzerlegung durchführen:

$${\color{red}{\int{\frac{1}{- k^{2} + r^{2}} d r}}} = {\color{red}{\int{\left(- \frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} + \frac{1}{2 \left(r - \left|{k}\right|\right) \left|{k}\right|}\right)d r}}}$$

Gliedweise integrieren:

$${\color{red}{\int{\left(- \frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} + \frac{1}{2 \left(r - \left|{k}\right|\right) \left|{k}\right|}\right)d r}}} = {\color{red}{\left(\int{\frac{1}{2 \left(r - \left|{k}\right|\right) \left|{k}\right|} d r} - \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$ mit $$$c=\frac{1}{2 \left|{k}\right|}$$$ und $$$f{\left(r \right)} = \frac{1}{- k + r}$$$ an:

$$- \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} + {\color{red}{\int{\frac{1}{2 \left(r - \left|{k}\right|\right) \left|{k}\right|} d r}}} = - \int{\frac{1}{2 \left(r + \left|{k}\right|\right) \left|{k}\right|} d r} + {\color{red}{\left(\frac{\int{\frac{1}{- k + r} d r}}{2 \left|{k}\right|}\right)}}$$

Sei $$$u=- k + r$$$.

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

Somit,

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

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

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

Zur Erinnerung: $$$u=- k + r$$$:

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

Wende die Konstantenfaktorregel $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$ mit $$$c=\frac{1}{2 \left|{k}\right|}$$$ und $$$f{\left(r \right)} = \frac{1}{k + r}$$$ an:

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

Sei $$$u=k + r$$$.

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

Das Integral lässt sich umschreiben als

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

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

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

Zur Erinnerung: $$$u=k + r$$$:

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \frac{1}{- k^{2} + r^{2}}\, dr = \frac{\ln\left(\left|{k - r}\right|\right) - \ln\left(\left|{k + r}\right|\right)}{2 \left|{k}\right|} + C$$$A