Question

4) Calcule a integral
$\int\ {} \frac{du}{u \sqrt{9u^2+4} } \,$

Answer 1

$I=\displaystyle\int\!\frac{du}{u\sqrt{9u^2+4}}\\\\\\ =\int\!\frac{1}{u\sqrt{9u^2+4}}\cdot \frac{18u}{18u}\,du\\\\\\ =\int\!\frac{18u\,du}{18u^2\sqrt{9u^2+4}}\\\\\\ =\int\!\frac{18u\,du}{2\cdot 9u^2\sqrt{9u^2+4}}\\\\\\ =\frac{1}{2}\int\!\frac{18u\,du}{9u^2\sqrt{9u^2+4}}~~~~~\mathbf{(i)}$


Façamos a seguinte substituição:

$\sqrt{9u^2+4}=w\\\\ 9u^2+4=w^2~~\Rightarrow~~\left\{ \!\begin{array}{l} 18u\,du=2w\,dw\\\\ 9u^2=w^2-4 \end{array} \right.$


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

$=\displaystyle\frac{1}{2}\int\!\frac{2w\,dw}{(w^2-4)\cdot w}\\\\\\ =\int\!\frac{dw}{w^2-4}\\\\\\ =\int\!\frac{dw}{(w-2)(w+2)}~~~~~~\mathbf{(ii)}$

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Em $\mathbf{(ii)},$ a função a ser integrada é uma função racional na variável $w,$ que pode ser decomposta em frações parciais:

$\dfrac{1}{(w-2)(w+2)}=\dfrac{A}{w-2}+\dfrac{B}{w+2}$


Multiplicando os dois lados acima por $(w-2)(w+2),$ ficamos com

$1=\dfrac{A(w-2)(w+2)}{w-2}+\dfrac{B(w-2)(w+2)}{w+2}\\\\\\ 1=A(w+2)+B(w-2)\\\\ 1=Aw+2A+Bw-2B\\\\ 1=(A+B)w+2A-2B\\\\0w+1=(A+B)w+2A-2B~~~~~~\mathbf{(iii)}$


Por identidade polinomial em $\mathbf{(iii)},$ tiramos que

$\left\{ \begin{array}{l} A+B=0\\\\ 2A-2B=1 \end{array}\right.$


Resolvendo o sistema acima para $A$ e $B,$ encontramos

$A=\dfrac{1}{4}~~\text{ e }~~B=-\,\dfrac{1}{4}$

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Portanto, a integral $\mathbf{(ii)}$ fica

$=\displaystyle\int\!\left(\frac{\frac{1}{4}}{w-2}-\frac{\frac{1}{4}}{w+2} \right )dw\\\\\\ =\frac{1}{4}\int\!\frac{1}{w-2}\,dw-\frac{1}{4}\int\!\frac{1}{w+2}\,dw\\\\\\ =\frac{1}{4}\,\mathrm{\ell n}\,|w-2|-\frac{1}{4}\,\mathrm{\ell n}\,|w+2|+C\\\\\\ =\frac{1}{4}\,\mathrm{\ell n}\big|\sqrt{9u^2+4}-2\big|-\frac{1}{4}\,\mathrm{\ell n}\big|\sqrt{9u^2+4}+2\big|+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{du}{u\sqrt{9u^2+4}}=\frac{1}{4}\,\mathrm{\ell n}\big|\sqrt{9u^2+4}-2\big|-\frac{1}{4}\,\mathrm{\ell n}\big|\sqrt{9u^2+4}+2\big|+C \end{array}}$


Bons estudos! :-)