Integral de $$$x^{3} e^{4 x^{2}}$$$

Encontre $$$\int x^{3} e^{4 x^{2}}\, dx$$$.

Seja $$$u=x^{2}$$$.

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

Assim,

$${\color{red}{\int{x^{3} e^{4 x^{2}} d x}}} = {\color{red}{\int{\frac{u e^{4 u}}{2} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = u e^{4 u}$$$:

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

Para a integral $$$\int{u e^{4 u} d u}$$$, use integração por partes $$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$.

Sejam $$$\operatorname{c}=u$$$ e $$$\operatorname{dv}=e^{4 u} du$$$.

Então $$$\operatorname{dc}=\left(u\right)^{\prime }du=1 du$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{4 u} d u}=\frac{e^{4 u}}{4}$$$ (os passos podem ser vistos »).

Logo,

$$\frac{{\color{red}{\int{u e^{4 u} d u}}}}{2}=\frac{{\color{red}{\left(u \cdot \frac{e^{4 u}}{4}-\int{\frac{e^{4 u}}{4} \cdot 1 d u}\right)}}}{2}=\frac{{\color{red}{\left(\frac{u e^{4 u}}{4} - \int{\frac{e^{4 u}}{4} d u}\right)}}}{2}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{4}$$$ e $$$f{\left(u \right)} = e^{4 u}$$$:

$$\frac{u e^{4 u}}{8} - \frac{{\color{red}{\int{\frac{e^{4 u}}{4} d u}}}}{2} = \frac{u e^{4 u}}{8} - \frac{{\color{red}{\left(\frac{\int{e^{4 u} d u}}{4}\right)}}}{2}$$

Seja $$$v=4 u$$$.

Então $$$dv=\left(4 u\right)^{\prime }du = 4 du$$$ (veja os passos »), e obtemos $$$du = \frac{dv}{4}$$$.

Assim,

$$\frac{u e^{4 u}}{8} - \frac{{\color{red}{\int{e^{4 u} d u}}}}{8} = \frac{u e^{4 u}}{8} - \frac{{\color{red}{\int{\frac{e^{v}}{4} d v}}}}{8}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ usando $$$c=\frac{1}{4}$$$ e $$$f{\left(v \right)} = e^{v}$$$:

$$\frac{u e^{4 u}}{8} - \frac{{\color{red}{\int{\frac{e^{v}}{4} d v}}}}{8} = \frac{u e^{4 u}}{8} - \frac{{\color{red}{\left(\frac{\int{e^{v} d v}}{4}\right)}}}{8}$$

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

$$\frac{u e^{4 u}}{8} - \frac{{\color{red}{\int{e^{v} d v}}}}{32} = \frac{u e^{4 u}}{8} - \frac{{\color{red}{e^{v}}}}{32}$$

Recorde que $$$v=4 u$$$:

$$\frac{u e^{4 u}}{8} - \frac{e^{{\color{red}{v}}}}{32} = \frac{u e^{4 u}}{8} - \frac{e^{{\color{red}{\left(4 u\right)}}}}{32}$$

Recorde que $$$u=x^{2}$$$:

$$- \frac{e^{4 {\color{red}{u}}}}{32} + \frac{{\color{red}{u}} e^{4 {\color{red}{u}}}}{8} = - \frac{e^{4 {\color{red}{x^{2}}}}}{32} + \frac{{\color{red}{x^{2}}} e^{4 {\color{red}{x^{2}}}}}{8}$$

Portanto,

$$\int{x^{3} e^{4 x^{2}} d x} = \frac{x^{2} e^{4 x^{2}}}{8} - \frac{e^{4 x^{2}}}{32}$$

Simplifique:

$$\int{x^{3} e^{4 x^{2}} d x} = \frac{\left(4 x^{2} - 1\right) e^{4 x^{2}}}{32}$$

Adicione a constante de integração:

$$\int{x^{3} e^{4 x^{2}} d x} = \frac{\left(4 x^{2} - 1\right) e^{4 x^{2}}}{32}+C$$

$$$\int x^{3} e^{4 x^{2}}\, dx = \frac{\left(4 x^{2} - 1\right) e^{4 x^{2}}}{32} + C$$$A