Integral de $$$2^{\sqrt{x}}$$$

Encontre $$$\int 2^{\sqrt{x}}\, dx$$$.

Alterar a base:

$${\color{red}{\int{2^{\sqrt{x}} d x}}} = {\color{red}{\int{e^{\sqrt{x} \ln{\left(2 \right)}} d x}}}$$

Seja $$$u=\sqrt{x} \ln{\left(2 \right)}$$$.

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

Assim,

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

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

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

Para a integral $$$\int{u e^{u} d u}$$$, use integração por partes $$$\int \operatorname{\theta} \operatorname{dv} = \operatorname{\theta}\operatorname{v} - \int \operatorname{v} \operatorname{d\theta}$$$.

Sejam $$$\operatorname{\theta}=u$$$ e $$$\operatorname{dv}=e^{u} du$$$.

Então $$$\operatorname{d\theta}=\left(u\right)^{\prime }du=1 du$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (os passos podem ser vistos »).

A integral torna-se

$$\frac{2 {\color{red}{\int{u e^{u} d u}}}}{\ln{\left(2 \right)}^{2}}=\frac{2 {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}}{\ln{\left(2 \right)}^{2}}=\frac{2 {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}}{\ln{\left(2 \right)}^{2}}$$

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

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

Recorde que $$$u=\sqrt{x} \ln{\left(2 \right)}$$$:

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

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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