Integral de $$$\frac{e^{2 x}}{e^{x} + 1}$$$

Halla $$$\int \frac{e^{2 x}}{e^{x} + 1}\, dx$$$.

Sea $$$u=e^{x}$$$.

Entonces $$$du=\left(e^{x}\right)^{\prime }dx = e^{x} dx$$$ (los pasos pueden verse »), y obtenemos que $$$e^{x} dx = du$$$.

Por lo tanto,

$${\color{red}{\int{\frac{e^{2 x}}{e^{x} + 1} d x}}} = {\color{red}{\int{\frac{u}{u + 1} d u}}}$$

Reescribe y separa la fracción:

$${\color{red}{\int{\frac{u}{u + 1} d u}}} = {\color{red}{\int{\left(1 - \frac{1}{u + 1}\right)d u}}}$$

Integra término a término:

$${\color{red}{\int{\left(1 - \frac{1}{u + 1}\right)d u}}} = {\color{red}{\left(\int{1 d u} - \int{\frac{1}{u + 1} d u}\right)}}$$

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

$$- \int{\frac{1}{u + 1} d u} + {\color{red}{\int{1 d u}}} = - \int{\frac{1}{u + 1} d u} + {\color{red}{u}}$$

Sea $$$v=u + 1$$$.

Entonces $$$dv=\left(u + 1\right)^{\prime }du = 1 du$$$ (los pasos pueden verse »), y obtenemos que $$$du = dv$$$.

Por lo tanto,

$$u - {\color{red}{\int{\frac{1}{u + 1} d u}}} = u - {\color{red}{\int{\frac{1}{v} d v}}}$$

La integral de $$$\frac{1}{v}$$$ es $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$u - {\color{red}{\int{\frac{1}{v} d v}}} = u - {\color{red}{\ln{\left(\left|{v}\right| \right)}}}$$

Recordemos que $$$v=u + 1$$$:

$$u - \ln{\left(\left|{{\color{red}{v}}}\right| \right)} = u - \ln{\left(\left|{{\color{red}{\left(u + 1\right)}}}\right| \right)}$$

Recordemos que $$$u=e^{x}$$$:

$$- \ln{\left(\left|{1 + {\color{red}{u}}}\right| \right)} + {\color{red}{u}} = - \ln{\left(\left|{1 + {\color{red}{e^{x}}}}\right| \right)} + {\color{red}{e^{x}}}$$

Por lo tanto,

$$\int{\frac{e^{2 x}}{e^{x} + 1} d x} = e^{x} - \ln{\left(e^{x} + 1 \right)}$$

Añade la constante de integración:

$$\int{\frac{e^{2 x}}{e^{x} + 1} d x} = e^{x} - \ln{\left(e^{x} + 1 \right)}+C$$

$$$\int \frac{e^{2 x}}{e^{x} + 1}\, dx = \left(e^{x} - \ln\left(e^{x} + 1\right)\right) + C$$$A