Sr Examen
Integral de x/(x^3-x^2-x-1) dx
Solución
Ha introducido
[src]
1 / | | x | --------------- dx | 3 2 | x - x - x - 1 | / 0
$$\int\limits_{0}^{1} \frac{x}{\left(- x + \left(x^{3} - x^{2}\right)\right) - 1}\, dx$$
Integral(x/(x^3 - x^2 - x - 1), (x, 0, 1))
Respuesta (Indefinida)
[src]
/ | / / 2 \\ | x | 3 |1 220*t 22*t|| | --------------- dx = C + RootSum|44*t - 2*t - 1, t -> t*log|-- + x - ------ - ----|| | 3 2 \ \17 17 17 // | x - x - x - 1 | /
$$\int \frac{x}{\left(- x + \left(x^{3} - x^{2}\right)\right) - 1}\, dx = C + \operatorname{RootSum} {\left(44 t^{3} - 2 t - 1, \left( t \mapsto t \log{\left(- \frac{220 t^{2}}{17} - \frac{22 t}{17} + x + \frac{1}{17} \right)} \right)\right)}$$
Respuesta
[src]
/ / 2 \\ / / 2 \\
| 3 |1 220*t 22*t|| | 3 |18 220*t 22*t||
- RootSum|44*t - 2*t - 1, t -> t*log|-- - ------ - ----|| + RootSum|44*t - 2*t - 1, t -> t*log|-- - ------ - ----||
\ \17 17 17 // \ \17 17 17 //
$$- \operatorname{RootSum} {\left(44 t^{3} - 2 t - 1, \left( t \mapsto t \log{\left(- \frac{220 t^{2}}{17} - \frac{22 t}{17} + \frac{1}{17} \right)} \right)\right)} + \operatorname{RootSum} {\left(44 t^{3} - 2 t - 1, \left( t \mapsto t \log{\left(- \frac{220 t^{2}}{17} - \frac{22 t}{17} + \frac{18}{17} \right)} \right)\right)}$$
=
=
/ / 2 \\ / / 2 \\
| 3 |1 220*t 22*t|| | 3 |18 220*t 22*t||
- RootSum|44*t - 2*t - 1, t -> t*log|-- - ------ - ----|| + RootSum|44*t - 2*t - 1, t -> t*log|-- - ------ - ----||
\ \17 17 17 // \ \17 17 17 //
$$- \operatorname{RootSum} {\left(44 t^{3} - 2 t - 1, \left( t \mapsto t \log{\left(- \frac{220 t^{2}}{17} - \frac{22 t}{17} + \frac{1}{17} \right)} \right)\right)} + \operatorname{RootSum} {\left(44 t^{3} - 2 t - 1, \left( t \mapsto t \log{\left(- \frac{220 t^{2}}{17} - \frac{22 t}{17} + \frac{18}{17} \right)} \right)\right)}$$
-RootSum(44*_t^3 - 2*_t - 1, Lambda(_t, _t*log(1/17 - 220*_t^2/17 - 22*_t/17))) + RootSum(44*_t^3 - 2*_t - 1, Lambda(_t, _t*log(18/17 - 220*_t^2/17 - 22*_t/17)))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.