Integral von $$$\frac{1}{- a + t}$$$ nach $$$t$$$

Bestimme $$$\int \frac{1}{- a + t}\, dt$$$.

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

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

Also,

$${\color{red}{\int{\frac{1}{- a + t} d t}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \frac{1}{- a + t}\, dt = \ln\left(\left|{a - t}\right|\right) + C$$$A