Question

1) Calcule a integral

$\int\ {} \frac{dx}{x^3 \sqrt{x^2-9} } \,$

Answer 1

$\displaystyle\int\!\frac{dx}{x^3\sqrt{x^2-9}}\\\\\\ =\int\!\frac{dx}{x^3\sqrt{x^2-9}}\cdot \frac{2x}{2x}\,dx\\\\\\ =\int\!\frac{2x\,dx}{2x^4\sqrt{x^2-9}}~~~~~~\mathbf{(i)}$


Faça a seguinte substituição:

$\sqrt{x^2-9}=u~~~~(u\ge 0)\\\\ x^2-9=u^2~~\Rightarrow~~2x\,dx=2u\,du\\\\ x^2=u^2+9\\\\ x^4=(u^2+9)^2$


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

$=\displaystyle\int\!\dfrac{2u\,du}{2(u^2+9)^2\cdot u}\\\\\\ =\int\!\dfrac{du}{(u^2+9)^2}~~~~~~\mathbf{(ii)}$


Faça agora a seguinte substituição (trigonométrica)

$u=3\,\mathrm{tg\,}t~~\Rightarrow~~du=3\sec^2 t\,dt\\\\ u^2+9=(3\,\mathrm{tg\,}t)^2+9\\\\ u^2+9=9\,\mathrm{tg^2\,}t+9\\\\ u^2+9=9\,(\mathrm{tg^2\,}t+1)\\\\ u^2+9=9\sec^2 t$


Substituindo em $\mathbf{(ii)},$ temos

$=\displaystyle\int\!\frac{3\sec^2 t\,dt}{(9\sec^2 t)^2}\\\\\\ \int\!\frac{3\sec^2 t\,dt}{81\sec^4 t}\\\\\\=\frac{1}{27}\int\!\frac{dt}{\sec^2 t}\\\\\\ =\frac{1}{27}\int\!\cos^2 t\,dt\\\\\\ =\frac{1}{27}\int\!\frac{1}{2}\left(1+\cos 2t\right )dt\\\\\\ =\frac{1}{27}\int\!\left(\frac{1}{2}+\frac{1}{2}\cos 2t\right )dt$

$=\dfrac{1}{27}\cdot \left(\dfrac{t}{2}+\dfrac{1}{4}\,\mathrm{sen\,}2t \right )+C\\\\\\ =\dfrac{1}{27}\cdot \left(\frac{t}{2}+\dfrac{1}{2}\,\mathrm{sen\,}t\cos t \right )+C\\\\\\ =\dfrac{t}{54}+\dfrac{1}{54}\,\mathrm{sen\,}t\cos t+C$


Agora, como voltar para a variável original $x?$

Observe o triângulo retângulo em anexo:

Dele tiramos que

$\left\{ \!\begin{array}{l} \mathrm{sen\,}t=\dfrac{u}{\sqrt{u^2+9}}\\\\ \cos t=\dfrac{3}{\sqrt{u^2+9}}\\\\ t=\mathrm{arctg}\left(\dfrac{u}{3}\right) \end{array} \right.$


Substituindo em $\mathbf{(iii)}$ ficamos com

$=\dfrac{1}{54}\mathrm{arctg}\left(\dfrac{u}{3} \right )+\dfrac{1}{54}\cdot\dfrac{u}{\sqrt{u^2+9}}\cdot \dfrac{3}{\sqrt{u^2+9}}+C\\\\\\ =\dfrac{1}{54}\mathrm{arctg}\left(\dfrac{u}{3} \right )+\dfrac{1}{18}\cdot\dfrac{u}{u^2+9}+C\\\\\\ =\dfrac{1}{54}\,\mathrm{arctg}\left(\dfrac{\sqrt{x^2-9}}{3} \right )+\dfrac{1}{18}\cdot\dfrac{\sqrt{x^2-9}}{x^2}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{dx}{x^3\sqrt{x^2-9}}=\frac{1}{54}\,\mathrm{arctg}\left(\dfrac{\sqrt{x^2-9}}{3} \right )+\dfrac{1}{18}\cdot\dfrac{\sqrt{x^2-9}}{x^2}+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \frac{dx}{x^3\sqrt{x^2-9}}\\\\\\=\int \frac{1}{27\sec ^2\left(u\right)}du\\\\\\=\frac{1}{27}\cdot \int \frac{1}{\sec ^2\left(u\right)}du\\\\\\\frac{1}{27}\cdot \int \frac{1+\cos \left(2u\right)}{2}du\\\\\\=\frac{1}{27}\cdot \frac{1}{2}\cdot \int \:1+\cos \left(2u\right)du\\\\\\\frac{1}{27}\cdot \frac{1}{2}\left(\int \:1du+\int \cos \left(2u\right)du\right)\\\\\\=\frac{1}{27}\cdot \frac{1}{2}\left(u+\frac{1}{2}\sin \left(2u\right)\right)$

$\sf \displaystyle =\frac{1}{27}\cdot \frac{1}{2}\left(\arcsec \left(\frac{1}{3}x\right)+\frac{1}{2}\sin \left(2\arcsec \left(\frac{1}{3}x\right)\right)\right)\\\\\\\frac{1}{54}\left(\arcsec \left(\frac{1}{3}x\right)+\frac{1}{2}\sin \left(2\arcsec \left(\frac{1}{3}x\right)\right)\right)\\\\\\\to \boxed{\sf \frac{1}{54}\left(\arcsec \left(\frac{1}{3}x\right)+\frac{1}{2}\sin \left(2\arcsec \left(\frac{1}{3}x\right)\right)\right)+C}$