Question

integral de 1 a mais infinito 1/(3x+3)^^2? heeelllpppp :)

Answer 1

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Calcular a integral imprópria

     $\mathsf{\displaystyle\int_1^\infty\!\frac{1}{(3x+3)^2}\,dx}$


Encontrando a primitiva:

     $\mathsf{\displaystyle\int\!\frac{1}{(3x+3)^2}\,dx}\\\\\\ =\mathsf{\displaystyle\int\!\frac{1}{\big[3(x+1)\big]^2}\,dx}\\\\\\ =\mathsf{\displaystyle\int\!\frac{1}{9(x+1)^2}\,dx}\\\\\\ =\mathsf{\displaystyle\frac{1}{9}\int\!(x+1)^{-2}\,dx}$


Faça uma substituição de variável:

     $\mathsf{x+1=u\quad\Rightarrow\quad dx=du}$


e a integral fica

     $=\mathsf{\displaystyle\frac{1}{9}\int\!u^{-2}\,du}\\\\\\ =\mathsf{\dfrac{1}{9}\cdot \dfrac{u^{-2+1}}{-2+1}+C}\\\\\\ =\mathsf{\dfrac{1}{9}\cdot \dfrac{u^{-1}}{-1}+C}\\\\\\ =\mathsf{\dfrac{1}{9}\cdot \left(-\,\dfrac{1}{u}\right)+C}\\\\\\ =\mathsf{-\,\dfrac{1}{9u}+C}\\\\\\ =\mathsf{-\,\dfrac{1}{9(x+1)}+C}$


Computando a integral imprópria, temos que

     $\mathsf{\displaystyle\int_1^\infty\!\frac{1}{(3x+3)^2}\,dx}\\\\\\ =\mathsf{\dfrac{1}{9(x+1)}\bigg|_1^\infty}\\\\\\ =\mathsf{\underset{x\to \infty}{\ell im}\left[-\,\dfrac{1}{9(x+1)}\right]-\left[-\,\dfrac{1}{9(1+1)}\right]}\\\\\\ =\mathsf{0+\dfrac{1}{9\cdot 2}}$

     $=\mathsf{\dfrac{1}{18}}$   <————   esta é a resposta.


Bons estudos! :-)