Question

Qual a integral de x^2*e^x

Answer 1

$I=\displaystyle\int\!x^2\,e^x\,dx$


Método de integração por partes:

$\begin{array}{lcl} u=x^2&~\Rightarrow~&du=2x\,dx\\\\ dv=e^x\,dx&~\Leftarrow~&v=e^x \end{array}\\\\\\\\ \displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \int\!x^2\,e^x\,dx=x^2\,e^x-\int\!e^x\cdot 2x\,dx\\\\\\ I=x^2\,e^x-\underbrace{\int\!2x\,e^x\,dx}_{I_1}~~~~~~\mathbf{(i)}$

______________

Calculando $I_1$ também por partes:

$\begin{array}{lcl} u=2x&~\Rightarrow~&du=2\,dx\\\\ dv=e^x\,dx&~\Leftarrow~&v=e^x \end{array}\\\\\\\\ \displaystyle\int\!u\,dv=uv-\int\!v\,du\\\\\\ \int\!2x\,e^x\,dx=2x\,e^x-\int\!e^x\cdot 2\,dx\\\\\\ I_1=2x\,e^x-2\int\!e^x\,dx\\\\\\ I_1=2x\,e^x-2e^x$


Substituindo em $\mathbf{(i)},$ ficamos com

$I=x^2\,e^x-(2x\,e^x-2e^x)+C\\\\ I=x^2\,e^x-2x\,e^x+2e^x+C\\\\ I=(x^2-2x+2)\,e^x+C\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!x^2\,e^x\,dx=(x^2-2x+2)\,e^x+C \end{array}}$


Bons estudos! :-)