Sr Examen
Integral de sqrt(1+1/(x^4)) dx
Solución
Ha introducido
[src]
1 / | | ________ | / 1 | / 1 + -- dx | / 4 | \/ x | / 0
$$\int\limits_{0}^{1} \sqrt{1 + \frac{1}{x^{4}}}\, dx$$
Integral(sqrt(1 + 1/(x^4)), (x, 0, 1))
Respuesta (Indefinida)
[src]
_ / | pi*I\
/ |_ |-1/2, -1/4 | e |
| x*Gamma(-1/4)* | | | -----|
| ________ 2 1 | 3/4 | 4 |
| / 1 \ | x /
| / 1 + -- dx = C - ---------------------------------------
| / 4 4*Gamma(3/4)
| \/ x
|
/
$$\int \sqrt{1 + \frac{1}{x^{4}}}\, dx = C - \frac{x \Gamma\left(- \frac{1}{4}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle| {\frac{e^{i \pi}}{x^{4}}} \right)}}{4 \Gamma\left(\frac{3}{4}\right)}$$
Gráfica
Respuesta
[src]
_
|_ /-1/2, -1/4 | \
-Gamma(-1/4)* | | | -1|
2 1 \ 3/4 | /
------------------------------------
4*Gamma(3/4)
$$- \frac{\Gamma\left(- \frac{1}{4}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle| {-1} \right)}}{4 \Gamma\left(\frac{3}{4}\right)}$$
=
=
_
|_ /-1/2, -1/4 | \
-Gamma(-1/4)* | | | -1|
2 1 \ 3/4 | /
------------------------------------
4*Gamma(3/4)
$$- \frac{\Gamma\left(- \frac{1}{4}\right) {{}_{2}F_{1}\left(\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle| {-1} \right)}}{4 \Gamma\left(\frac{3}{4}\right)}$$
-gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), -1)/(4*gamma(3/4))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.