Integraal van $$$\frac{2 x^{3}}{\sqrt{5 - 8 x^{4}}}$$$

Bepaal $$$\int \frac{2 x^{3}}{\sqrt{5 - 8 x^{4}}}\, dx$$$.

Zij $$$u=5 - 8 x^{4}$$$.

Dan $$$du=\left(5 - 8 x^{4}\right)^{\prime }dx = - 32 x^{3} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$x^{3} dx = - \frac{du}{32}$$$.

Dus,

$${\color{red}{\int{\frac{2 x^{3}}{\sqrt{5 - 8 x^{4}}} d x}}} = {\color{red}{\int{\left(- \frac{1}{16 \sqrt{u}}\right)d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=- \frac{1}{16}$$$ en $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

$${\color{red}{\int{\left(- \frac{1}{16 \sqrt{u}}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{\sqrt{u}} d u}}{16}\right)}}$$

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=- \frac{1}{2}$$$:

$$- \frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{16}=- \frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{16}=- \frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{16}=- \frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{16}=- \frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{16}$$

We herinneren eraan dat $$$u=5 - 8 x^{4}$$$:

$$- \frac{\sqrt{{\color{red}{u}}}}{8} = - \frac{\sqrt{{\color{red}{\left(5 - 8 x^{4}\right)}}}}{8}$$

Dus,

$$\int{\frac{2 x^{3}}{\sqrt{5 - 8 x^{4}}} d x} = - \frac{\sqrt{5 - 8 x^{4}}}{8}$$

Voeg de integratieconstante toe:

$$\int{\frac{2 x^{3}}{\sqrt{5 - 8 x^{4}}} d x} = - \frac{\sqrt{5 - 8 x^{4}}}{8}+C$$

$$$\int \frac{2 x^{3}}{\sqrt{5 - 8 x^{4}}}\, dx = - \frac{\sqrt{5 - 8 x^{4}}}{8} + C$$$A