Integral de (x^3-1)^(1/2) dx
Solución
Respuesta (Indefinida)
[src]
/ _
| |_ /-1/2, 1/3 | 3\
| ________ I*x*Gamma(1/3)* | | | x |
| / 3 2 1 \ 4/3 | /
| \/ x - 1 dx = C + ------------------------------------
| 3*Gamma(4/3)
/
$$\int \sqrt{x^{3} - 1}\, dx = C + \frac{i x \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {x^{3}} \right)}}{3 \Gamma\left(\frac{4}{3}\right)}$$
_ _
|_ /-1/2, 1/3 | \ |_ /-1/2, 1/3 | \
I*Gamma(1/3)* | | | 1| 2*I*Gamma(1/3)* | | | 8|
2 1 \ 4/3 | / 2 1 \ 4/3 | /
- --------------------------------- + -----------------------------------
3*Gamma(4/3) 3*Gamma(4/3)
$$- \frac{i \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {1} \right)}}{3 \Gamma\left(\frac{4}{3}\right)} + \frac{2 i \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {8} \right)}}{3 \Gamma\left(\frac{4}{3}\right)}$$
=
_ _
|_ /-1/2, 1/3 | \ |_ /-1/2, 1/3 | \
I*Gamma(1/3)* | | | 1| 2*I*Gamma(1/3)* | | | 8|
2 1 \ 4/3 | / 2 1 \ 4/3 | /
- --------------------------------- + -----------------------------------
3*Gamma(4/3) 3*Gamma(4/3)
$$- \frac{i \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {1} \right)}}{3 \Gamma\left(\frac{4}{3}\right)} + \frac{2 i \Gamma\left(\frac{1}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle| {8} \right)}}{3 \Gamma\left(\frac{4}{3}\right)}$$
-i*gamma(1/3)*hyper((-1/2, 1/3), (4/3,), 1)/(3*gamma(4/3)) + 2*i*gamma(1/3)*hyper((-1/2, 1/3), (4/3,), 8)/(3*gamma(4/3))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.