Question

Gostaria que resolvessem essa integral, ∫$\frac{1}{u^{3}*\sqrt{u^{2}-9 } }$ du
desde ja obrigado.

Answer 1

$\int \frac{du}{u^{3}\sqrt{u^{2}-9}}$

Faça

$u=3\sec{\theta}\\du=3\sec(\theta).\tan(\theta) d\theta$

$\int \frac{du}{u^{3}\sqrt{u^{2}-9}}du = \int \frac{3 \sec( \theta). \tan( \theta)d\theta}{27 { \sec }^{3}\theta.3 \tan(\theta) }$

$\frac{1}{27} \int { \cos}^{2}\theta \: d\theta$

$\frac{1}{27} . \frac{1}{2} \int1 + \cos(2\theta)d\theta \\ = \frac{1}{54}\theta + \frac{1}{54.\cancel{2}}.\cancel{2}. \sin(\theta). \cos(\theta) \\ + k$

$= \frac{1}{54} (\theta + \sin(\theta). \cos(\theta)) + k$

$\frac{1}{54}(arc \sec( \frac{u}{3}) + \frac{ \sqrt{ {u}^{2} - 9 } }{u} . \frac{3}{u} ) + k$

$\int \frac{du}{u^{3}\sqrt{u^{2}-9}}du \\ = \frac{1}{54}(arc \sec( \frac{u}{3}) + \frac{ 3\sqrt{ {u}^{2} - 9 } }{ {u}^{2} } )+ k$

Answer 2

Calcular a integral indefinida

    $\mathsf{\displaystyle\int\frac{1}{u^3\sqrt{u^2-9}}\,du}$

Multiplique o numerador e o denominador do integrando por 2u:

    $\mathsf{=\displaystyle\int\frac{2u}{u^3\sqrt{u^2-9}\cdot 2u}\,du}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\frac{1}{u^4\sqrt{u^2-9}}\cdot 2u\,du\qquad (i)}$

Faça a seguinte substituição racionalizante:

    $\mathsf{v=\sqrt{u^2-9}}\\\\ \mathsf{\Longrightarrow\quad v^2=u^2-9}\\\\ \mathsf{\Longrightarrow\quad u^2=v^2+9\quad\Longrightarrow\quad 2u\,du=2v\,dv}\\\\ \mathsf{\Longrightarrow\quad u^4=(v^2+9)^2}$

Substituindo, a integral fica:

    $\mathsf{=\displaystyle\frac{1}{2}\int\frac{1}{(v^2+9)^2\cdot \diagup\!\!\!\! v}\cdot 2\diagup\!\!\!\! v\,dv}\\\\\\ \mathsf{=\displaystyle\frac{1}{2}\int\frac{1}{(v^2+9)^2}\cdot 2\,dv}\\\\\\ \mathsf{=\displaystyle\int\frac{1}{(v^2+9)^2}\,dv}\\\\\\ \mathsf{=\displaystyle\int\frac{1}{(v^2+3^2)^2}\,dv\qquad (ii)}$

Agora temos que integrar uma função racional em v. Você pode resolvê-la por frações parciais. Mas nesse caso, como o denominador envolve uma soma de quadrados e já não mais raízes quadradas, fica mais fácil fazer uma substituição trigonométrica:

    $\mathsf{v=3\,tg\,\theta\quad\Longrightarrow\quad dv=3\,sec^2\,\theta\,d\theta}\\\\\\\ \mathsf{\Longrightarrow\quad v^2+3^2=(3\,tg\,\theta)^2+3^2}\\\\ \mathsf{\Longleftrightarrow\quad v^2+3^2=3^2\,tg^2\,\theta+3^2}\\\\ \mathsf{\Longleftrightarrow\quad v^2+3^2=3^2\,(tg^2\,\theta+1)}\\\\ \mathsf{\Longleftrightarrow\quad v^2+3^2=3^2\,sec^2\,\theta}\\\\ \mathsf{\Longrightarrow\quad (v^2+3^2)^2=(3^2\,sec^2\,\theta)^2}\\\\ \mathsf{\Longleftrightarrow\quad (v^2+3^2)^2=3^4\,sec^4\,\theta}$

Substituindo em (ii), a integral fica

    $\mathsf{=\displaystyle\int\frac{1}{3^4\,sec^4\,\theta}\cdot 3\,sec^2\,\theta\,d\theta}\\\\\\ \mathsf{=\displaystyle\frac{1}{3^3}\int\frac{1}{sec^2\,\theta}\,d\theta}\\\\\\ \mathsf{=\displaystyle\frac{1}{3^3}\int cos^2\,\theta\,d\theta}\\\\\\ \mathsf{=\displaystyle\frac{1}{3^3}\int \dfrac{1+cos\,2\theta}{2} \,d\theta}\\\\\\ \mathsf{=\displaystyle\frac{1}{3^3\cdot 2}\int (1+cos\,2\theta)\,d\theta}\\\\\\ \mathsf{=\displaystyle\frac{1}{3^3\cdot 2}\int d\theta+\frac{1}{3^3\cdot 2}\int cos\,2\theta\,d\theta}$

    $\mathsf{=\displaystyle\frac{1}{3^3\cdot 2}\int d\theta+\frac{1}{3^3\cdot 2^2}\int cos\,2\theta\cdot 2\,d\theta}\\\\\\ \mathsf{=\dfrac{1}{3^3\cdot 2}\cdot \theta+\dfrac{1}{3^3\cdot 2^2}\cdot sen\,2\theta+C}\\\\\\ \mathsf{=\dfrac{1}{3^3\cdot 2}\cdot \theta+\dfrac{1}{3^3\cdot 2^2}\cdot 2\,sen\,\theta\,cos\,\theta+C}\\\\\\ \mathsf{=\dfrac{1}{3^3\cdot 2}\cdot \theta+\dfrac{1}{3^3\cdot 2}\cdot sen\,\theta\,cos\,\theta+C\qquad (iii)}$

Substituindo θ de volta para v:

    $\mathsf{v=3\,tg\,\theta\quad\Longrightarrow\quad \theta=arctg\!\left(\dfrac{v}{3} \right),~~com~-\,\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}}\\\\\\ \mathsf{\Longrightarrow\quad tg\,\theta=\dfrac{v}{3}}\\\\\\ \mathsf{\Longleftrightarrow\quad \dfrac{sen\,\theta}{cos\,\theta}=\dfrac{v}{3}}\\\\\\ \mathsf{\Longleftrightarrow\quad 3\,sen\,\theta=v\,cos\,\theta}\\\\ \mathsf{\Longrightarrow\quad (3\,sen\,\theta)^2=(v\,cos\,\theta)^2}\\\\ \mathsf{\Longleftrightarrow\quad 3^2\,sen^2\,\theta=v^2\,cos^2\,\theta}\\\\ \mathsf{\Longleftrightarrow\quad 3^2\,(1-cos^2\,\theta)=v^2\,cos^2\,\theta}$

    $\mathsf{\Longleftrightarrow\quad 3^2-3^2\,cos^2\,\theta=v^2\,cos^2\,\theta}\\\\ \mathsf{\Longleftrightarrow\quad 3^2=v^2\,cos^2\,\theta+3^2\,cos^2\,\theta}\\\\ \mathsf{\Longleftrightarrow\quad 3^2=(v^2+3^2)\,cos^2\,\theta}\\\\ \mathsf{\Longleftrightarrow\quad cos^2\,\theta=\dfrac{3^2}{v^2+3^2}}\\\\\\ \mathsf{\Longleftrightarrow\quad cos\,\theta=\pm\,\dfrac{3}{\sqrt{v^2+3^2}}}$

Como $\mathsf{-\,\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2},}$  segue que cos θ é positivo. Portanto,

    $\mathsf{\Longrightarrow\quad cos\,\theta=\dfrac{3}{\sqrt{v^2+3^2}}\qquad \checkmark}$

Pela definição de tangente, devemos ter então

    $\mathsf{\Longrightarrow\quad sen\,\theta=\dfrac{v}{\sqrt{v^2+3^2}}\qquad \checkmark}$

Subsituindo em (iii), obtemos

    $\mathsf{=\dfrac{1}{3^3\cdot 2}\,arctg\!\left(\dfrac{v}{3}\right)+\dfrac{1}{3^3\cdot 2}\cdot \dfrac{v}{\sqrt{v^2+3^2}}\cdot \dfrac{3}{\sqrt{v^2+3^2}}+C}\\\\\\ \mathsf{=\dfrac{1}{54}\,arctg\!\left(\dfrac{v}{3}\right)+\dfrac{1}{18}\cdot \dfrac{v}{v^2+3^2}+C}$

Substituindo de volta para a variável u, sendo $\mathsf{v=\sqrt{u^2-9},}$  finalmente encontramos

    $\mathsf{=\dfrac{1}{54}\,arctg\bigg(\dfrac{\sqrt{u^2-9}}{3}\bigg)+\dfrac{\sqrt{u^2-9}}{18u^2}+C\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)