Question

Calcule a integral da função racional :
∫ 16x+4/x^2-4

Answer 1

Boa tarde Camila!

Solução!

Sendo a integral.

$\displaystyle \int\left (\frac{16x+4}{ x^{2} -4} \right ) dx\\\\\\\ Vamos~~ reescrever~~ a~~ integral ~~com~~ o~~ denominador~~fatorado.\\\\\\\\ \displaystyle \int\left (\frac{16x+4}{ (x-2) \times(x+2)} \right ) dx\\\\\\\ Somas~~de~~fracoes~~parciais.$


$\left (\dfrac{16x+4}{ (x-2) \times(x+2)} )\right =\left ( \dfrac{A}{(x-2)}+ \dfrac{B}{(x+2)} \right ) \\\\\\\ \left ( \dfrac{A}{(x-2)}+ \dfrac{B}{(x+2)} \right )=16x+4\\\\\\\ A(x+2)+B(x-2)=16x+4\\\\\\\ Encontramos~~um ~~sistema~~linear,fazendo~~a~~ distributiva.\\\\\\\ (Ax+2A)+(Bx-2B)=16x+4\\\\\\ Agrupando~~os~~termos~~semelhantes.\\\\\\ (Ax+Bx+2A-2B)=16x+4\\\\\\ \begin{cases}Ax+Bx=16x\\ 2A-2B=4\end{cases} \\\\\\\\ Cancelando~~a~~variavel~~x~~fica~~assim.$

$\begin{cases}A+B=16\\ 2A-2B=4\end{cases} \\\\\\\\ Vamos~~dividir~~a~~segunda~~equacao~~por~~2.\\\\\ \begin{cases}A+B=16\\ 2A-2B=4:(2)\end{cases} \\\\\\\\ \begin{cases}A+B=16\\ A-B=2\end{cases} \\\\\\\\ Resolvendo~~o~~sistema.\\\\\\ A+A=16+2\\\\\ 2A=18\\\\ A= \dfrac{18}{2}\\\\\ A=9\\\\\ Vamos ~~agora~~ determinar~~ o~~ valor~~ de~~ B.\\\\\\ A+B=16\\\\ 9+B=16\\\\\ B=16-9\\\\\ B=7$

$Retomando~~a~~integral.\\\\\\\\\ \displaystyle \int\left (\frac{16x+4}{ x^{2} -4} \right ) dx=\displaystyle \int\left (\frac{16x+4}{ (x-2) \times(x+2)} \right )dx$


$\displaystyle \int\left ( \dfrac{A}{(x-2)}+ \dfrac{B}{(x+2)} \right )dx=\displaystyle \int\left ( \dfrac{9}{(x-2)}+ \dfrac{7}{(x+2)} \right ) dx$


$\displaystyle \int\left ( \dfrac{9}{(x-2)}+ \dfrac{7}{(x+2)} \right ) dx=\displaystyle \int\left ( \dfrac{9}{(x-2)}\right )dx +\displaystyle \int\left (\dfrac{7}{(x+2)} \right )dx$



$\displaystyle \int\left ( \dfrac{9}{(x-2)}\right )dx +\displaystyle \int\left (\dfrac{7}{(x+2)} \right ) dx = 9ln|x-2|+7ln|x+2|+c$


Portanto!

$\boxed{Resposta:\displaystyle \int\left (\frac{16x+4}{ x^{2} -4} \right )dx=9ln|x-2|+7ln|x+2|+c }$

Boa tarde!
Bons estudos!