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

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

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

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

Logo,

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

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 $$$f{\left(u \right)} = e^{u^{2}}$$$:

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

Esta integral (Função erro imaginária) não possui forma fechada:

$$\frac{{\color{red}{\int{e^{u^{2}} d u}}}}{2} = \frac{{\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erfi}{\left(u \right)}}{2}\right)}}}{2}$$

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

$$\frac{\sqrt{\pi} \operatorname{erfi}{\left({\color{red}{u}} \right)}}{4} = \frac{\sqrt{\pi} \operatorname{erfi}{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Portanto,

$$\int{e^{4 x^{2}} d x} = \frac{\sqrt{\pi} \operatorname{erfi}{\left(2 x \right)}}{4}$$

Adicione a constante de integração:

$$\int{e^{4 x^{2}} d x} = \frac{\sqrt{\pi} \operatorname{erfi}{\left(2 x \right)}}{4}+C$$

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