Question

2) Calcule a integral

$\int\ {} \frac{dx}{ \sqrt{x^2-25} } \,$

Answer 1

$I=\displaystyle\int\!\frac{dx}{\sqrt{x^2-25}}~~~~~~\mathbf{(i)}$


Faça a seguinte substituição (trigonométrica)

$x=5\sec t~~~~(0<t<\pi/2)~~\Rightarrow~~dx=5\sec t\,\mathrm{tg\,}t\,dt\\\\\\ x^2-25=(5\sec t)^2-25\\\\ x^2-25=25\sec^2 t-25\\\\ x^2-25=25\cdot (\sec^2 t-1)\\\\ x^2-25=25\,\mathrm{tg^2\,}t\\\\ \sqrt{x^2-25}=5\,\mathrm{tg\,}t~~~~(\text{pois }0<t<\pi/2)$


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

$=\displaystyle\int\!\frac{5\sec t\,\mathrm{tg\,}t\,dt}{5\,\mathrm{tg\,}t}\\\\\\ =\int\!\sec t\,dt\\\\\\ =\mathrm{\ell n}\left|\sec t +\mathrm{tg\,}t\right|+C_1~~~~~~\mathbf{(ii)}$


Para voltar à variável $x,$ basta observar as substituições feitas inicialmente:

$\left\{\! \begin{array}{l} \mathrm{\sec\,}t=\dfrac{x}{5}\\\\ \mathrm{tg\,}t=\dfrac{\sqrt{x^2-25}}{5} \end{array} \right.$


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

$=\mathrm{\ell n}\left|\dfrac{x}{5} +\dfrac{\sqrt{x^2-25}}{5}\right|+C_1\\\\\\ =\mathrm{\ell n}\left|\dfrac{x+\sqrt{x^2-25}}{5}\right|+C_1\\\\\\ =\mathrm{\ell n}\big|x+\sqrt{x^2-25}\big|-\mathrm{\ell n\,}5+C_1\\\\\\ =\mathrm{\ell n}\big|x+\sqrt{x^2-25}\big|+C\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{dx}{\sqrt{x^2-25}}=\mathrm{\ell n}\big|x+\sqrt{x^2-25}\big|+C \end{array}}$


Obs.: Esta primitiva foi obtida apenas considerando $x>5.$ Mas, se considerarmos $x<-5$ e fizermos a substituição trigonométrica, chegamos à mesma expressão acima.


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \frac{dx}{\sqrt{x^2-25}}\\\\\\=\int \sec \left(u\right)du\\\\\\=\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|\\\\\\=\ln \left|\tan \left(\arcsec \left(\frac{1}{5}x\right)\right)+\sec \left(\arcsec \left(\frac{1}{5}x\right)\right)\right|\\\\\\=\ln \left(\frac{1}{5}\left|\sqrt{x^2-25}+x\right|\right)\\\\\\\boxed{\sf =\ln \left(\frac{1}{5}\left|\sqrt{x^2-25}+x\right|\right)+C}$