Integral de $$$e^{x} \cosh{\left(x \right)}$$$

Encontre $$$\int e^{x} \cosh{\left(x \right)}\, dx$$$.

Reescreva a função hiperbólica em termos da função exponencial:

$${\color{red}{\int{e^{x} \cosh{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{e^{x}}{2} + \frac{e^{- x}}{2}\right) e^{x} d x}}}$$

Simplifique o integrando:

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(x \right)} = \left(e^{x} + e^{- x}\right) e^{x}$$$:

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

Simplify:

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

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=1$$$:

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

Seja $$$u=2 x$$$.

Então $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{2}$$$.

A integral torna-se

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = e^{u}$$$:

$$\frac{x}{2} + \frac{{\color{red}{\int{\frac{e^{u}}{2} d u}}}}{2} = \frac{x}{2} + \frac{{\color{red}{\left(\frac{\int{e^{u} d u}}{2}\right)}}}{2}$$

A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:

$$\frac{x}{2} + \frac{{\color{red}{\int{e^{u} d u}}}}{4} = \frac{x}{2} + \frac{{\color{red}{e^{u}}}}{4}$$

Recorde que $$$u=2 x$$$:

$$\frac{x}{2} + \frac{e^{{\color{red}{u}}}}{4} = \frac{x}{2} + \frac{e^{{\color{red}{\left(2 x\right)}}}}{4}$$

Portanto,

$$\int{e^{x} \cosh{\left(x \right)} d x} = \frac{x}{2} + \frac{e^{2 x}}{4}$$

Adicione a constante de integração:

$$\int{e^{x} \cosh{\left(x \right)} d x} = \frac{x}{2} + \frac{e^{2 x}}{4}+C$$

$$$\int e^{x} \cosh{\left(x \right)}\, dx = \left(\frac{x}{2} + \frac{e^{2 x}}{4}\right) + C$$$A