Integral von $$$\frac{t^{2}}{t^{2} - 2}$$$

Bestimme $$$\int \frac{t^{2}}{t^{2} - 2}\, dt$$$.

Forme den Bruch um und zerlege ihn:

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

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, dt = c t$$$ mit $$$c=1$$$ an:

$$\int{\frac{2}{t^{2} - 2} d t} + {\color{red}{\int{1 d t}}} = \int{\frac{2}{t^{2} - 2} d t} + {\color{red}{t}}$$

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

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

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

$$t + 2 {\color{red}{\int{\frac{1}{t^{2} - 2} d t}}} = t + 2 {\color{red}{\int{\left(- \frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} + \frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)}\right)d t}}}$$

Gliedweise integrieren:

$$t + 2 {\color{red}{\int{\left(- \frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} + \frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)}\right)d t}}} = t + 2 {\color{red}{\left(\int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \int{\frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} d t}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=\frac{\sqrt{2}}{4}$$$ und $$$f{\left(t \right)} = \frac{1}{t + \sqrt{2}}$$$ an:

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - 2 {\color{red}{\int{\frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} d t}}} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - 2 {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{t + \sqrt{2}} d t}}{4}\right)}}$$

Sei $$$u=t + \sqrt{2}$$$.

Dann $$$du=\left(t + \sqrt{2}\right)^{\prime }dt = 1 dt$$$ (die Schritte sind » zu sehen), und es gilt $$$dt = du$$$.

Also,

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{t + \sqrt{2}} d t}}}}{2} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{2} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Zur Erinnerung: $$$u=t + \sqrt{2}$$$:

$$t - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} = t - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(t + \sqrt{2}\right)}}}\right| \right)}}{2} + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=\frac{\sqrt{2}}{4}$$$ und $$$f{\left(t \right)} = \frac{1}{t - \sqrt{2}}$$$ an:

$$t - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2} + 2 {\color{red}{\int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t}}} = t - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2} + 2 {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{t - \sqrt{2}} d t}}{4}\right)}}$$

Sei $$$u=t - \sqrt{2}$$$.

Dann $$$du=\left(t - \sqrt{2}\right)^{\prime }dt = 1 dt$$$ (die Schritte sind » zu sehen), und es gilt $$$dt = du$$$.

Das Integral wird zu

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

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

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

Zur Erinnerung: $$$u=t - \sqrt{2}$$$:

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

Daher,

$$\int{\frac{t^{2}}{t^{2} - 2} d t} = t + \frac{\sqrt{2} \ln{\left(\left|{t - \sqrt{2}}\right| \right)}}{2} - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{t^{2}}{t^{2} - 2} d t} = t + \frac{\sqrt{2} \ln{\left(\left|{t - \sqrt{2}}\right| \right)}}{2} - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2}+C$$

$$$\int \frac{t^{2}}{t^{2} - 2}\, dt = \left(t + \frac{\sqrt{2} \ln\left(\left|{t - \sqrt{2}}\right|\right)}{2} - \frac{\sqrt{2} \ln\left(\left|{t + \sqrt{2}}\right|\right)}{2}\right) + C$$$A