Integral von $$$\frac{1}{\left(4 - x\right)^{\frac{3}{2}}}$$$

Bestimme $$$\int \frac{1}{\left(4 - x\right)^{\frac{3}{2}}}\, dx$$$.

Sei $$$u=4 - x$$$.

Dann $$$du=\left(4 - x\right)^{\prime }dx = - dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = - du$$$.

Daher,

$${\color{red}{\int{\frac{1}{\left(4 - x\right)^{\frac{3}{2}}} d x}}} = {\color{red}{\int{\left(- \frac{1}{u^{\frac{3}{2}}}\right)d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = \frac{1}{u^{\frac{3}{2}}}$$$ an:

$${\color{red}{\int{\left(- \frac{1}{u^{\frac{3}{2}}}\right)d u}}} = {\color{red}{\left(- \int{\frac{1}{u^{\frac{3}{2}}} d u}\right)}}$$

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=- \frac{3}{2}$$$ an:

$$- {\color{red}{\int{\frac{1}{u^{\frac{3}{2}}} d u}}}=- {\color{red}{\int{u^{- \frac{3}{2}} d u}}}=- {\color{red}{\frac{u^{- \frac{3}{2} + 1}}{- \frac{3}{2} + 1}}}=- {\color{red}{\left(- 2 u^{- \frac{1}{2}}\right)}}=- {\color{red}{\left(- \frac{2}{\sqrt{u}}\right)}}$$

Zur Erinnerung: $$$u=4 - x$$$:

$$2 \frac{1}{\sqrt{{\color{red}{u}}}} = 2 \frac{1}{\sqrt{{\color{red}{\left(4 - x\right)}}}}$$

Daher,

$$\int{\frac{1}{\left(4 - x\right)^{\frac{3}{2}}} d x} = \frac{2}{\sqrt{4 - x}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\left(4 - x\right)^{\frac{3}{2}}} d x} = \frac{2}{\sqrt{4 - x}}+C$$

$$$\int \frac{1}{\left(4 - x\right)^{\frac{3}{2}}}\, dx = \frac{2}{\sqrt{4 - x}} + C$$$A