Sr Examen
Integral de -e^(3*x)/(2*(e^(2*x)+1)) dx
Solución
Ha introducido
[src]
1 / | | 3*x | -E | ------------ dx | / 2*x \ | 2*\E + 1/ | / 0
$$\int\limits_{0}^{1} \frac{\left(-1\right) e^{3 x}}{2 \left(e^{2 x} + 1\right)}\, dx$$
Integral((-E^(3*x))/((2*(E^(2*x) + 1))), (x, 0, 1))
Respuesta (Indefinida)
[src]
/ | | 3*x x | -E e / 2 / x\\ | ------------ dx = C - -- + RootSum\16*z + 1, i -> i*log\4*i + e // | / 2*x \ 2 | 2*\E + 1/ | /
$$\int \frac{\left(-1\right) e^{3 x}}{2 \left(e^{2 x} + 1\right)}\, dx = C - \frac{e^{x}}{2} + \operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(4 i + e^{x} \right)} \right)\right)}$$
Gráfica
Respuesta
[src]
1 / 2 \ E / 2 \ - - RootSum\16*z + 1, i -> i*log(1 + 4*i)/ - - + RootSum\16*z + 1, i -> i*log(E + 4*i)/ 2 2
$$- \operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(4 i + 1 \right)} \right)\right)} + \operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(4 i + e \right)} \right)\right)} - \frac{e}{2} + \frac{1}{2}$$
=
=
1 / 2 \ E / 2 \ - - RootSum\16*z + 1, i -> i*log(1 + 4*i)/ - - + RootSum\16*z + 1, i -> i*log(E + 4*i)/ 2 2
$$- \operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(4 i + 1 \right)} \right)\right)} + \operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(4 i + e \right)} \right)\right)} - \frac{e}{2} + \frac{1}{2}$$
1/2 - RootSum(16*_z^2 + 1, Lambda(_i, _i*log(1 + 4*_i))) - E/2 + RootSum(16*_z^2 + 1, Lambda(_i, _i*log(E + 4*_i)))
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.