[tex]\Large\boxed{\begin{array}{l}\rm Quando~o~integrando~\acute e~da~forma\\\sf \sqrt{a^2+x^2}\\\sf use~x=a\,tg(\theta)~de~modo~que~\sqrt{a^2+x^2}=a\,sec(\theta)\\\sf e~dx=a\,sec^2(\theta)\,d\theta\end{array}}[/tex]
[tex]\Large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{dx}{\sqrt{4+x^2}}\\\underline{\rm fac_{\!\!,}a}\\\sf x=2\,tg(\theta)\\\sf\sqrt{4+x^2}=2\,sec(\theta)\\\sf dx=2\,sec^2(\theta)\,d\theta\\\displaystyle\sf\int\dfrac{dx}{\sqrt{4+x^2}}=\int\dfrac{\diagup\!\!\!2\,\diagdown\!\!\!\!\!\!sec^2(\theta)}{\diagup\!\!\!2\,\diagdown\!\!\!\!\!sec(\theta)}d\theta\\\displaystyle\sf =\int sec(\theta)d\theta=\ell n|sec(\theta)+tg(\theta)|+k\end{array}}[/tex]
[tex]\Large\boxed{\begin{array}{l}\underline{\rm Usando~o~tri\hat angulo~auxiliar~temos:}\\\sf sec(\theta)=\dfrac{\sqrt{4+x^2}}{2}~tg(\theta)=\dfrac{x}{2}\\\\\displaystyle\sf\int\dfrac{dx}{\sqrt{4+x^2}}=\ell n\bigg|\dfrac{\sqrt{4+x^2}+x}{2}\bigg|+k\end{array}}[/tex]
