Integral von $$$\frac{1}{\sqrt{v}}$$$

Bestimme $$$\int \frac{1}{\sqrt{v}}\, dv$$$.

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

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

Daher,

$$\int{\frac{1}{\sqrt{v}} d v} = 2 \sqrt{v}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sqrt{v}} d v} = 2 \sqrt{v}+C$$

$$$\int \frac{1}{\sqrt{v}}\, dv = 2 \sqrt{v} + C$$$A