Integralen av $$$4 \sqrt{x} - \frac{4}{\sqrt{x}}$$$

Bestäm $$$\int \left(4 \sqrt{x} - \frac{4}{\sqrt{x}}\right)\, dx$$$.

Integrera termvis:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=4$$$ och $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=- \frac{1}{2}$$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=4$$$ och $$$f{\left(x \right)} = \sqrt{x}$$$:

$$- 8 \sqrt{x} + {\color{red}{\int{4 \sqrt{x} d x}}} = - 8 \sqrt{x} + {\color{red}{\left(4 \int{\sqrt{x} d x}\right)}}$$

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=\frac{1}{2}$$$:

$$- 8 \sqrt{x} + 4 {\color{red}{\int{\sqrt{x} d x}}}=- 8 \sqrt{x} + 4 {\color{red}{\int{x^{\frac{1}{2}} d x}}}=- 8 \sqrt{x} + 4 {\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=- 8 \sqrt{x} + 4 {\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}$$

Alltså,

$$\int{\left(4 \sqrt{x} - \frac{4}{\sqrt{x}}\right)d x} = \frac{8 x^{\frac{3}{2}}}{3} - 8 \sqrt{x}$$

Förenkla:

$$\int{\left(4 \sqrt{x} - \frac{4}{\sqrt{x}}\right)d x} = \frac{8 \sqrt{x} \left(x - 3\right)}{3}$$

Lägg till integrationskonstanten:

$$\int{\left(4 \sqrt{x} - \frac{4}{\sqrt{x}}\right)d x} = \frac{8 \sqrt{x} \left(x - 3\right)}{3}+C$$

$$$\int \left(4 \sqrt{x} - \frac{4}{\sqrt{x}}\right)\, dx = \frac{8 \sqrt{x} \left(x - 3\right)}{3} + C$$$A