Sr Examen
Integral de (2^x)/((4^x)+4)dx dx
Solución
Ha introducido
[src]
1 / | | x | 2 | ------ dx | x | 4 + 4 | / 0
$$\int\limits_{0}^{1} \frac{2^{x}}{4^{x} + 4}\, dx$$
Integral(2^x/(4^x + 4), (x, 0, 1))
Respuesta (Indefinida)
[src]
/ / x\ | |2 | | x atan|--| | 2 \2 / | ------ dx = C + -------- | x 2*log(2) | 4 + 4 | /
$$\int \frac{2^{x}}{4^{x} + 4}\, dx = C + \frac{\operatorname{atan}{\left(\frac{2^{x}}{2} \right)}}{2 \log{\left(2 \right)}}$$
Gráfica
Respuesta
[src]
/ 2 \ / 2 \
RootSum\16*z + 1, i -> i*log(2 + 8*i)/ RootSum\16*z + 1, i -> i*log(1 + 8*i)/
--------------------------------------- - ---------------------------------------
log(2) log(2)
$$- \frac{\operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(8 i + 1 \right)} \right)\right)}}{\log{\left(2 \right)}} + \frac{\operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(8 i + 2 \right)} \right)\right)}}{\log{\left(2 \right)}}$$
=
=
/ 2 \ / 2 \
RootSum\16*z + 1, i -> i*log(2 + 8*i)/ RootSum\16*z + 1, i -> i*log(1 + 8*i)/
--------------------------------------- - ---------------------------------------
log(2) log(2)
$$- \frac{\operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(8 i + 1 \right)} \right)\right)}}{\log{\left(2 \right)}} + \frac{\operatorname{RootSum} {\left(16 z^{2} + 1, \left( i \mapsto i \log{\left(8 i + 2 \right)} \right)\right)}}{\log{\left(2 \right)}}$$
RootSum(16*_z^2 + 1, Lambda(_i, _i*log(2 + 8*_i)))/log(2) - RootSum(16*_z^2 + 1, Lambda(_i, _i*log(1 + 8*_i)))/log(2)
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.