Integral de $$$e^{\sqrt{33} \sqrt{x}}$$$
Calculadora relacionada: Calculadora de Integrais Definidas e Impróprias
Sua entrada
Encontre $$$\int e^{\sqrt{33} \sqrt{x}}\, dx$$$.
Solução
Seja $$$u=\sqrt{33} \sqrt{x}$$$.
Então $$$du=\left(\sqrt{33} \sqrt{x}\right)^{\prime }dx = \frac{\sqrt{33}}{2 \sqrt{x}} dx$$$ (veja os passos »), e obtemos $$$\frac{dx}{\sqrt{x}} = \frac{2 \sqrt{33} du}{33}$$$.
Portanto,
$${\color{red}{\int{e^{\sqrt{33} \sqrt{x}} d x}}} = {\color{red}{\int{\frac{2 u e^{u}}{33} d u}}}$$
Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{2}{33}$$$ e $$$f{\left(u \right)} = u e^{u}$$$:
$${\color{red}{\int{\frac{2 u e^{u}}{33} d u}}} = {\color{red}{\left(\frac{2 \int{u e^{u} d u}}{33}\right)}}$$
Para a integral $$$\int{u e^{u} d u}$$$, use integração por partes $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.
Sejam $$$\operatorname{g}=u$$$ e $$$\operatorname{dv}=e^{u} du$$$.
Então $$$\operatorname{dg}=\left(u\right)^{\prime }du=1 du$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{u} d u}=e^{u}$$$ (os passos podem ser vistos »).
Logo,
$$\frac{2 {\color{red}{\int{u e^{u} d u}}}}{33}=\frac{2 {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}}{33}=\frac{2 {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}}{33}$$
A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:
$$\frac{2 u e^{u}}{33} - \frac{2 {\color{red}{\int{e^{u} d u}}}}{33} = \frac{2 u e^{u}}{33} - \frac{2 {\color{red}{e^{u}}}}{33}$$
Recorde que $$$u=\sqrt{33} \sqrt{x}$$$:
$$- \frac{2 e^{{\color{red}{u}}}}{33} + \frac{2 {\color{red}{u}} e^{{\color{red}{u}}}}{33} = - \frac{2 e^{{\color{red}{\sqrt{33} \sqrt{x}}}}}{33} + \frac{2 {\color{red}{\sqrt{33} \sqrt{x}}} e^{{\color{red}{\sqrt{33} \sqrt{x}}}}}{33}$$
Portanto,
$$\int{e^{\sqrt{33} \sqrt{x}} d x} = \frac{2 \sqrt{33} \sqrt{x} e^{\sqrt{33} \sqrt{x}}}{33} - \frac{2 e^{\sqrt{33} \sqrt{x}}}{33}$$
Simplifique:
$$\int{e^{\sqrt{33} \sqrt{x}} d x} = \frac{2 \left(\sqrt{33} \sqrt{x} - 1\right) e^{\sqrt{33} \sqrt{x}}}{33}$$
Adicione a constante de integração:
$$\int{e^{\sqrt{33} \sqrt{x}} d x} = \frac{2 \left(\sqrt{33} \sqrt{x} - 1\right) e^{\sqrt{33} \sqrt{x}}}{33}+C$$
Resposta
$$$\int e^{\sqrt{33} \sqrt{x}}\, dx = \frac{2 \left(\sqrt{33} \sqrt{x} - 1\right) e^{\sqrt{33} \sqrt{x}}}{33} + C$$$A