Integral von $$$\frac{- x^{\frac{3}{2}} + 4 x^{5} - 3}{x^{2}}$$$

Bestimme $$$\int \frac{- x^{\frac{3}{2}} + 4 x^{5} - 3}{x^{2}}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\frac{- x^{\frac{3}{2}} + 4 x^{5} - 3}{x^{2}} d x}}} = {\color{red}{\int{\left(4 x^{3} - \frac{3}{x^{2}} - \frac{1}{\sqrt{x}}\right)d x}}}$$

Gliedweise integrieren:

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

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

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

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

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

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

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=4$$$ und $$$f{\left(x \right)} = x^{3}$$$ an:

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

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

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

Daher,

$$\int{\frac{- x^{\frac{3}{2}} + 4 x^{5} - 3}{x^{2}} d x} = - 2 \sqrt{x} + x^{4} + \frac{3}{x}$$

Vereinfachen:

$$\int{\frac{- x^{\frac{3}{2}} + 4 x^{5} - 3}{x^{2}} d x} = \frac{- 2 x^{\frac{3}{2}} + x^{5} + 3}{x}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{- x^{\frac{3}{2}} + 4 x^{5} - 3}{x^{2}} d x} = \frac{- 2 x^{\frac{3}{2}} + x^{5} + 3}{x}+C$$

$$$\int \frac{- x^{\frac{3}{2}} + 4 x^{5} - 3}{x^{2}}\, dx = \frac{- 2 x^{\frac{3}{2}} + x^{5} + 3}{x} + C$$$A