Integralen av $$$- 3 \sqrt{6} \sqrt{x} - x^{2} + x z^{2}$$$ med avseende på $$$x$$$

Bestäm $$$\int \left(- 3 \sqrt{6} \sqrt{x} - x^{2} + x z^{2}\right)\, dx$$$.

Integrera termvis:

$${\color{red}{\int{\left(- 3 \sqrt{6} \sqrt{x} - x^{2} + x z^{2}\right)d x}}} = {\color{red}{\left(- \int{x^{2} d x} - \int{3 \sqrt{6} \sqrt{x} d x} + \int{x z^{2} d x}\right)}}$$

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

$$- \int{3 \sqrt{6} \sqrt{x} d x} + \int{x z^{2} d x} - {\color{red}{\int{x^{2} d x}}}=- \int{3 \sqrt{6} \sqrt{x} d x} + \int{x z^{2} d x} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \int{3 \sqrt{6} \sqrt{x} d x} + \int{x z^{2} d x} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=z^{2}$$$ och $$$f{\left(x \right)} = x$$$:

$$- \frac{x^{3}}{3} - \int{3 \sqrt{6} \sqrt{x} d x} + {\color{red}{\int{x z^{2} d x}}} = - \frac{x^{3}}{3} - \int{3 \sqrt{6} \sqrt{x} d x} + {\color{red}{z^{2} \int{x d x}}}$$

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

$$- \frac{x^{3}}{3} + z^{2} {\color{red}{\int{x d x}}} - \int{3 \sqrt{6} \sqrt{x} d x}=- \frac{x^{3}}{3} + z^{2} {\color{red}{\frac{x^{1 + 1}}{1 + 1}}} - \int{3 \sqrt{6} \sqrt{x} d x}=- \frac{x^{3}}{3} + z^{2} {\color{red}{\left(\frac{x^{2}}{2}\right)}} - \int{3 \sqrt{6} \sqrt{x} d x}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=3 \sqrt{6}$$$ och $$$f{\left(x \right)} = \sqrt{x}$$$:

$$- \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2} - {\color{red}{\int{3 \sqrt{6} \sqrt{x} d x}}} = - \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2} - {\color{red}{\left(3 \sqrt{6} \int{\sqrt{x} d x}\right)}}$$

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

$$- \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2} - 3 \sqrt{6} {\color{red}{\int{\sqrt{x} d x}}}=- \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2} - 3 \sqrt{6} {\color{red}{\int{x^{\frac{1}{2}} d x}}}=- \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2} - 3 \sqrt{6} {\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=- \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2} - 3 \sqrt{6} {\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}$$

Alltså,

$$\int{\left(- 3 \sqrt{6} \sqrt{x} - x^{2} + x z^{2}\right)d x} = - 2 \sqrt{6} x^{\frac{3}{2}} - \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2}$$

Lägg till integrationskonstanten:

$$\int{\left(- 3 \sqrt{6} \sqrt{x} - x^{2} + x z^{2}\right)d x} = - 2 \sqrt{6} x^{\frac{3}{2}} - \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2}+C$$

$$$\int \left(- 3 \sqrt{6} \sqrt{x} - x^{2} + x z^{2}\right)\, dx = \left(- 2 \sqrt{6} x^{\frac{3}{2}} - \frac{x^{3}}{3} + \frac{x^{2} z^{2}}{2}\right) + C$$$A