Question

Calcule a integral por partes

Answer 1

Olá!
  
    Sejam   $u=x^2,\;\;dv=e^{2x}$  . Logo,


$\displaystyle \int{u\;dv}=uv-\displaystyle \int{v\;du} \\ \\ \displaystyle \int{x^2\cdot e^{2x}dx=x^2 \cdot \dfrac{e^{2x}}{2}- \displaystyle \int{\dfrac{e^{2x}}{2}\cdot 2x}\;dx = \\ \\ \\ = \dfrac{1}{2}x^2\cdot e^{2x}-\displaystyle\int{xe^{2x}}dx =$

$= \dfrac{1}{2}x^2\cdot e^{2x}-\left(x\cdot \dfrac{e^{2x}}{2}- \displaystyle\int{\dfrac{e^{2x}}{2}\cdot 1}dx\right) = \\ \\ \\ = \dfrac{1}{2}x^2\cdot e^{2x}-\left(\dfrac{1}{2}x\cdot e^{2x}-\dfrac{1}{2} \displaystyle\int{e^{2x}}dx\right) = \\ \\ \\ = \dfrac{1}{2}x^2\cdot e^{2x}-\dfrac{1}{2}x\cdot e^{2x}+\dfrac{1}{2}\cdot \dfrac{e^{2x}}{2}+k,\;\; k\in\mathbb{R} = \\ \\ = \dfrac{1}{2}\cdot e^{2x}\left(x^2-x+\dfrac{1}{2}\right)+k,\;\; k\in\mathbb{R}$

Bons estudos!

Answer 2

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$\displastyle\sf\int x^2e^{2x}~dx\\\sf fac_{\!\!,}a~u=x^2\implies du=2x~dx\\\sf dv=e^{2x}~dx\implies v=\dfrac{1}{2}e^{2x}\\\displaystyle\sf \int u\cdot dv=u\cdot v-\int v\cdot du\\\displaystyle\sf\int x^2e^{2x}~dx=x^2\cdot\dfrac{1}{2}e^{2x}-\int\dfrac{1}{\diagup\!\!\!2}e^{2x}\cdot\diagup\!\!\!2x~dx\\\displaystyle\sf\int x^2e^{2x}~dx=\dfrac{1}{2}x^2e^{2x}-\int x\cdot e^{2x}~dx\\\tt fac_{\!\!,}a~u_1=x\implies du_1=dx\\\sf dv_1=e^{2x}~dx\implies v_1=\dfrac{1}{2}e^{2x}$

$\displaystyle\sf\int u_1\cdot dv_1=u_1\cdot v_1-\int v_1~dv_1\\\displaystyle\sf\int x\cdot e^{2x}~dx=x\cdot\dfrac{1}{2}e^{2x}-\int\dfrac{1}{2}e^{2x}~dx\\\displastyle\sf\int x\cdot e^{2x}~dx=\dfrac{1}{2}x\cdot e^{2x}-\dfrac{1}{2}\cdot\dfrac{1}{2}e^{2x}=\dfrac{1}{2}e^{2x}\left(x-\dfrac{1}{2}\right)+c_1$

$\displaystyle\sf\int x^2\cdot e^{2x}~dx=\dfrac{1}{2}x^2\cdot e^{2x}-\dfrac{1}{2}e^{2x}\cdot\left(x-\dfrac{1}{2}\right)+k$