Integral von $$$\frac{1}{4 t^{2}}$$$

Bestimme $$$\int \frac{1}{4 t^{2}}\, dt$$$.

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

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

Wenden Sie die Potenzregel $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-2$$$ an:

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

Daher,

$$\int{\frac{1}{4 t^{2}} d t} = - \frac{1}{4 t}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{4 t^{2}} d t} = - \frac{1}{4 t}+C$$

$$$\int \frac{1}{4 t^{2}}\, dt = - \frac{1}{4 t} + C$$$A