Question

Calcule a seguinte integral:

$\displaystyle \int_{- \infty }^{+ \infty} \frac{x^2}{9+x^6} \, \, dx$

Elaborando de maneira bem exemplificada.

Answer 1

Avaliar a integral imprópria:

     $\displaystyle\int_{-\infty}^{\infty}\frac{x^2}{9+x^6}\,dx\\\\\\ =\int_{-\infty}^\infty\frac{1}{3}\cdot 3\cdot \frac{x^2}{9+(x^3)^2}\,dx\\\\\\ =\frac{1}{3}\int_{-\infty}^{\infty} \frac{1}{3^2+(x^3)^2}\cdot 3x^2\,dx$



Faça a seguinte substituição:

     $x^3=3\,\mathrm{tg\,}t\\\\ \Rightarrow\quad\left\{\!\begin{array}{l}t=\mathrm{arctg}\!\left(\dfrac{x^3}{3}\right)\\\\ 3x^2\,dx=3\sec^2 t\,dt \end{array}\right.$



Novos limites de integração em $t:$

     $\begin{array}{lcl}\textsf{Quando~~}x\to -\infty&\quad\Rightarrow\quad&t\to -\,\dfrac{\pi}{2} \end{array}\\\\\\ \begin{array}{lcl}\textsf{Quando~~}x\to \infty&\quad\Rightarrow\quad&t\to \dfrac{\pi}{2} \end{array}$



Então, a integral fica

     $\displaystyle=\frac{1}{3}\int_{-\pi/2}^{\pi/2}\,\frac{1}{3^2+(3\,\mathrm{tg\,}t)^2}\cdot 3\sec^2 t\,dt\\\\\\ =\frac{1}{3}\int_{-\pi/2}^{\pi/2}\,\frac{1}{3^2(1+\mathrm{tg^2\,}t)}\cdot 3\sec^2 t\,dt\\\\\\ =\frac{1}{3}\int_{-\pi/2}^{\pi/2}\,\frac{1}{3^2\sec^2 t}\cdot 3\sec^2 t\,dt\\\\\\ =\frac{1}{3}\int_{-\pi/2}^{\pi/2}\,\frac{1}{3}\,dt\\\\\\ =\frac{1}{9}\int_{-\pi/2}^{\pi/2}\,dt$

     $=\dfrac{1}{9}\cdot t\bigg|_{-\pi/2}^{\pi/2}\\\\\\ =\dfrac{1}{9}\cdot \left[\dfrac{\pi}{2}-\left(-\dfrac{\pi}{2}\right)\right]\\\\\\ =\dfrac{1}{9}\cdot \left[2\cdot\dfrac{\pi}{2}\right]$

     $=\dfrac{\pi}{9}\quad\longleftarrow\quad\textsf{resposta.}$



Bons estudos! :-)