Integral de $$$- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2$$$

Encontre $$$\int \left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)\, dx$$$.

Integre termo a termo:

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

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=2$$$:

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

Aplique a regra da potência $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=2$$$:

$$- 2 x + \int{e^{\sqrt{2} \sqrt{x}} d x} - {\color{red}{\int{x^{2} d x}}}=- 2 x + \int{e^{\sqrt{2} \sqrt{x}} d x} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- 2 x + \int{e^{\sqrt{2} \sqrt{x}} d x} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

Seja $$$u=\sqrt{2} \sqrt{x}$$$.

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

Logo,

$$- \frac{x^{3}}{3} - 2 x + {\color{red}{\int{e^{\sqrt{2} \sqrt{x}} d x}}} = - \frac{x^{3}}{3} - 2 x + {\color{red}{\int{u e^{u} d u}}}$$

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

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

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

Portanto,

$$- \frac{x^{3}}{3} - 2 x + {\color{red}{\int{u e^{u} d u}}}=- \frac{x^{3}}{3} - 2 x + {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}=- \frac{x^{3}}{3} - 2 x + {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}$$

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

$$u e^{u} - \frac{x^{3}}{3} - 2 x - {\color{red}{\int{e^{u} d u}}} = u e^{u} - \frac{x^{3}}{3} - 2 x - {\color{red}{e^{u}}}$$

Recorde que $$$u=\sqrt{2} \sqrt{x}$$$:

$$- \frac{x^{3}}{3} - 2 x - e^{{\color{red}{u}}} + {\color{red}{u}} e^{{\color{red}{u}}} = - \frac{x^{3}}{3} - 2 x - e^{{\color{red}{\sqrt{2} \sqrt{x}}}} + {\color{red}{\sqrt{2} \sqrt{x}}} e^{{\color{red}{\sqrt{2} \sqrt{x}}}}$$

Portanto,

$$\int{\left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)d x} = \sqrt{2} \sqrt{x} e^{\sqrt{2} \sqrt{x}} - \frac{x^{3}}{3} - 2 x - e^{\sqrt{2} \sqrt{x}}$$

Adicione a constante de integração:

$$\int{\left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)d x} = \sqrt{2} \sqrt{x} e^{\sqrt{2} \sqrt{x}} - \frac{x^{3}}{3} - 2 x - e^{\sqrt{2} \sqrt{x}}+C$$

$$$\int \left(- x^{2} + e^{\sqrt{2} \sqrt{x}} - 2\right)\, dx = \left(\sqrt{2} \sqrt{x} e^{\sqrt{2} \sqrt{x}} - \frac{x^{3}}{3} - 2 x - e^{\sqrt{2} \sqrt{x}}\right) + C$$$A