Question

5) Calcule a integral

$\int\ \frac{1}{x^2 \sqrt{x^2+4} } \, dx$

Answer 1

$I=\displaystyle\int\!\frac{1}{x^2\sqrt{x^2+4}}\,dx~~~~~~\mathbf{(i)}$


Faça a seguinte substituição trigonométrica:

$x=2\,\mathrm{tg\,}t~~\Rightarrow~~\left\{ \begin{array}{l} dx=2\sec^2 t\,dt\\\\ t=\mathrm{arctg}\left(\dfrac{x}{2} \right ) \end{array} \right.\\\\\\ x^2+4=(2\,\mathrm{tg\,}t)^2+4\\\\ x^2+4=4\,\mathrm{tg^2\,}t+4\\\\ x^2+4=4\cdot (\mathrm{tg^2\,}t+1)\\\\ x^2+4=4\sec^2 t\\\\ \sqrt{x^2+4}=\sqrt{4\sec^2 t}\\\\ \sqrt{x^2+4}=2\sec t~~~~(\text{pois }-\pi/2<t<0~~\text{ ou }~~0<t<\pi/2)$


Substituindo em $\mathbf{(i)},$ a integral fica

$=\displaystyle\int\!\frac{1}{(2\,\mathrm{tg\,}t)^2\cdot 2\sec t}\cdot 2\sec^2 t\,dt\\\\\\ =\int\!\frac{\diagup\!\!\!\! 2\sec t\cdot \sec t}{4\,\mathrm{tg^2\,}t\cdot \diagup\!\!\!\! 2\sec t}\,dt\\\\\\ =\frac{1}{4}\int\!\frac{\sec t}{\mathrm{tg^2\,}t}\,dt\\\\\\ =\frac{1}{4}\int\!\sec t\cdot \frac{1}{\mathrm{tg^2\,}t}\,dt\\\\\\ =\frac{1}{4}\int\!\frac{1}{\cos t}\cdot \frac{\cos^2 t}{\mathrm{sen^2\,}t}\,dt\\\\\\ =\frac{1}{4}\int\!\frac{\cos t}{\mathrm{sen^2\,}t}\,dt$

$=\displaystyle\frac{1}{4}\int\!\frac{1}{\mathrm{sen^2\,}t}\cdot \cos t\,dt\\\\\\ =\frac{1}{4}\int\!\frac{1}{u^2}\,du~~~~~~(u=\mathrm{sen\,}t)\\\\\\ =\frac{1}{4}\int\!u^{-2}\,du\\\\\\ =\frac{1}{4}\cdot \frac{u^{-2+1}}{-2+1}+C\\\\\\ =\frac{1}{4}\cdot \frac{u^{-1}}{-1}+C\\\\\\ =-\,\frac{1}{4}\cdot \frac{1}{u}+C\\\\\\ =-\,\frac{1}{4}\cdot \frac{1}{\mathrm{sen\,}t}+C\\\\\\ =-\,\frac{1}{4}\,\mathrm{cossec\,}t+C~~~~~~\mathbf{(ii)}$


Para voltar à variável original $x,$ tomamos como auxílio o triângulo retângulo em anexo. Daí, tiramos que

$\mathrm{cossec\,}t=\dfrac{\sqrt{x^2+4}}{x}$


e voltando a $\mathbf{(iii)},$ finalmente chegamos a

$=-\,\dfrac{1}{4}\cdot \dfrac{\sqrt{x^2+4}}{x}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{1}{x^2\sqrt{x^2+4}}\,dx=-\,\dfrac{\sqrt{x^2+4}}{4x}+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \frac{1}{x^2\sqrt{x^2+4}}dx\\\\\\=\int \frac{\sec \left(u\right)}{4\tan ^2\left(u\right)}du\\\\\\=\frac{1}{4}\cdot \int \frac{\sec \left(u\right)}{\tan ^2\left(u\right)}du\\\\\\=\frac{1}{4}\cdot \int \frac{\cos \left(u\right)}{\sin ^2\left(u\right)}du\\\\\\=\frac{1}{4}\cdot \int \frac{1}{v^2}dv\\\\\\=\frac{1}{4}\cdot \int \:v^{-2}dv\\\\\\\frac{1}{4}\cdot \frac{v^{-2+1}}{-2+1}$

$\sf \displaystyle$$\sf \displaystyle =\frac{1}{4}\cdot \frac{\sin ^{-2+1}\left(\arctan \left(\frac{1}{2}x\right)\right)}{-2+1}\\\\\\=-\frac{\sqrt{4+x^2}}{4x}\\\\\\\to \boxed{\sf -\frac{\sqrt{4+x^2}}{4x}+C}$