Question

Integral de (x^2+2x+1)e^7xdx

Answer 1

Resposta:   $\displaystyle\int (x^2+2x+1)\cdot e^{7x}\,dx=\frac{1}{343}\,(49x^2+84x+37)\cdot e^{7x}+C.$

Explicação passo a passo:

Calcular a integral indefinida:

    $\displaystyle\int (x^2+2x+1)\cdot e^{7x}\,dx$

Vamos integrar por partes:

    $\begin{array}{lcl} u=x^2+2x+1 &\quad\Longrightarrow\quad& du=(2x+2)\,dx=2(x+1)\,dx\\\\ dv=e^{7x}\,dx &\quad\Longleftarrow\quad& v=\dfrac{1}{7}\,e^{7x}\\\\ \end{array}$

    $\displaystyle\int u\,dv=uv-\int v\,du$

    $\displaystyle\Longrightarrow\quad\int (x^2+2x+1)\cdot e^{7x}\,dx=(x^2+2x+1)\cdot \frac{1}{7}\,e^{7x}-\int\frac{1}{7}\,e^{7x}\cdot 2(x+1)\,dx$

    $\displaystyle=\frac{1}{7}(x^2+2x+1)\cdot e^{7x}-\frac{2}{7}\int(x+1)\cdot e^{7x}\,dx\qquad \mathrm{(i)}$

Calculando por partes novamente:

    $\begin{array}{lcl} u=x+1 &\quad\Longrightarrow\quad& du=dx\\\\ dv=e^{7x}\,dx &\quad\Longleftarrow\quad& v=\dfrac{1}{7}\,e^{7x}\\\\ \end{array}$

    $\displaystyle\int u\,dv=uv-\int v\,du$

    $\displaystyle\Longrightarrow\quad\int (x+1)\cdot e^{7x}\,dx=(x+1)\cdot \frac{1}{7}\,e^{7x}-\int \frac{1}{7}\,e^{7x}\,dx\\\\\\=\frac{1}{7}\,(x+1)\cdot e^{7x}-\frac{1}{7}\int e^{7x}\,dx\\\\\\=\frac{1}{7}\,(x+1)\cdot e^{7x}-\frac{1}{7}\cdot \left(\frac{1}{7}\,e^{7x}\right)+C_1$

    $=\dfrac{1}{7}\,(x+1)\cdot e^{7x}-\dfrac{1}{49}\,e^{7x}+C_1\qquad\mathrm{(ii)}$

Substituindo em (i), obtemos

    $\displaystyle\int (x^2+2x+1)\cdot e^{7x}\,dx=\frac{1}{7}\,(x^2+2x+1)\cdot e^{7x}-\frac{2}{7}\cdot \left[\dfrac{1}{7}\,(x+1)\cdot e^{7x}-\dfrac{1}{49}\,e^{7x}\right]+C$

    $\displaystyle=\frac{1}{7}\,(x^2+2x+1)\cdot e^{7x}-\frac{2}{49}\,(x+1)\cdot e^{7x}+\dfrac{2}{343}\,e^{7x}+C\quad\longleftarrow\quad\mathsf{resposta.}$

Poderíamos parar por aqui, mas ainda dá para simplificar a expressão acima, reduzindo as frações ao mesmo denominador comum e colocando $e^{7x}$ em evidência:

    $=\dfrac{49}{343}\,(x^2+2x+1)\cdot e^{7x}-\dfrac{14}{343}\,(x+1)\cdot e^{7x}+\dfrac{2}{343}\,e^{7x}+C\\\\\\=\dfrac{1}{343}\,(49x^2+98x+49)\cdot e^{7x}-\dfrac{1}{343}\,(14x+14)\cdot e^{7x}+\dfrac{2}{343}\,e^{7x}+C$

    $=\dfrac{1}{343}\big[(49x^2+98x+49)-(14x+14)+2\big]\cdot e^{7x}+C$

Reduza os termos semelhantes do polinômio que está nos colchetes, e finalmente chegamos a:

    $=\dfrac{1}{343}\,(49x^2+84x+37)\cdot e^{7x}+C\quad\longleftarrow\quad\mathsf{resposta.}$

Dúvidas? Comente.

Bons estudos!