Integral de √(x^3-4) dx
Solución
Respuesta (Indefinida)
[src]
/ _ / | 3\
| |_ |-1/2, 1/3 | x |
| ________ 2*I*x*Gamma(1/3)* | | | --|
| / 3 2 1 \ 4/3 | 4 /
| \/ x - 4 dx = C + --------------------------------------
| 3*Gamma(4/3)
/
$$\int \sqrt{x^{3} - 4}\, dx = C + \frac{2 i x \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {\frac{x^{3}}{4}} \right)}}{3 \Gamma\left(\frac{4}{3}\right)}$$
_
|_ /-1/2, 1/3 | \
2*I*Gamma(1/3)* | | | 1/4|
2 1 \ 4/3 | /
-------------------------------------
3*Gamma(4/3)
$$\frac{2 i \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {\frac{1}{4}} \right)}}{3 \Gamma\left(\frac{4}{3}\right)}$$
=
_
|_ /-1/2, 1/3 | \
2*I*Gamma(1/3)* | | | 1/4|
2 1 \ 4/3 | /
-------------------------------------
3*Gamma(4/3)
$$\frac{2 i \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {\frac{1}{4}} \right)}}{3 \Gamma\left(\frac{4}{3}\right)}$$
2*i*gamma(1/3)*hyper((-1/2, 1/3), (4/3,), 1/4)/(3*gamma(4/3))
(0.0 + 1.93504509561959j)
(0.0 + 1.93504509561959j)
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.