Integral de $$$\frac{2 x^{3}}{\sqrt{5 - 6 x^{4}}}$$$

Halla $$$\int \frac{2 x^{3}}{\sqrt{5 - 6 x^{4}}}\, dx$$$.

Sea $$$u=5 - 6 x^{4}$$$.

Entonces $$$du=\left(5 - 6 x^{4}\right)^{\prime }dx = - 24 x^{3} dx$$$ (los pasos pueden verse »), y obtenemos que $$$x^{3} dx = - \frac{du}{24}$$$.

Por lo tanto,

$${\color{red}{\int{\frac{2 x^{3}}{\sqrt{5 - 6 x^{4}}} d x}}} = {\color{red}{\int{\left(- \frac{1}{12 \sqrt{u}}\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{12}$$$ y $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

$${\color{red}{\int{\left(- \frac{1}{12 \sqrt{u}}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{\sqrt{u}} d u}}{12}\right)}}$$

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=- \frac{1}{2}$$$:

$$- \frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{12}=- \frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{12}=- \frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{12}=- \frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{12}=- \frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{12}$$

Recordemos que $$$u=5 - 6 x^{4}$$$:

$$- \frac{\sqrt{{\color{red}{u}}}}{6} = - \frac{\sqrt{{\color{red}{\left(5 - 6 x^{4}\right)}}}}{6}$$

Por lo tanto,

$$\int{\frac{2 x^{3}}{\sqrt{5 - 6 x^{4}}} d x} = - \frac{\sqrt{5 - 6 x^{4}}}{6}$$

Añade la constante de integración:

$$\int{\frac{2 x^{3}}{\sqrt{5 - 6 x^{4}}} d x} = - \frac{\sqrt{5 - 6 x^{4}}}{6}+C$$

$$$\int \frac{2 x^{3}}{\sqrt{5 - 6 x^{4}}}\, dx = - \frac{\sqrt{5 - 6 x^{4}}}{6} + C$$$A