Question

Investigar a seguinte integral imprópria:

$\mathsf{\displaystyle\int\limits_{e}^{+\infty}\dfrac{dx}{x(lnx)^2}}}$


Resoluções sem cálculos e trolls serão eliminadas.

Answer 1

Avaliar a integral imprópria:

    $\mathsf{\displaystyle\int _e^{+\infty} \frac{dx}{x(\ell n\,x)^2}}\\\\\\ \mathsf{\displaystyle=\int _e^{+\infty} \frac{1}{(\ell n\,x)^2}\cdot \frac{1}{x}\,dx}$

Vamos fazer uma mudança de variável

    $\mathsf{u=\ell n\,x\quad\Longrightarrow \quad du=\dfrac{1}{x}\,dx}$

Novos limites de integração:

    $\begin{array}{lcl}\mathsf{Quando~~x=e}&\quad\Longrightarrow\quad&\mathsf{u=\ell n\,e}\\\\ &&\mathsf{u=1}\\\\\\ \mathsf{Quando~~x\to +\infty}&\quad\Longrightarrow\quad&\mathsf{u\to +\infty}\end{array}$

e a integral fica

    $\mathsf{\displaystyle=\int_1^{+\infty} \frac{1}{u^2}\,du}\\\\\\ \mathsf{\displaystyle=\int_1^{+\infty} u^{-2}\,du}\\\\\\ \mathsf{=\dfrac{u^{-2+1}}{-2+1}\bigg|_1^{+\infty}}\\\\\\ \mathsf{=\dfrac{u^{-1}}{-1}\bigg|_1^{+\infty}}\\\\\\ \mathsf{=-\,\dfrac{1}{u}\bigg|_1^{+\infty}}$

    $\mathsf{=\underset{u\to +\infty}{\ell im} \Big(\!-\dfrac{1}{u}\Big)-\Big(\!-\dfrac{1}{1}\Big)}\\\\\\ \mathsf{=0-(-1)}\\\\\\ \mathsf{=1\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)

Answer 2

Explicação passo-a-passo:

Cálculo da integral:

$\mathsf{\displaystyle\int\limts_{e}^{+\infty}~\mathsf{\dfrac{dx}{x(lnx)^2}} }$

Seja:

$\mathsf{lnx~=~u}$

$\mathsf{du~=~\dfrac{1}{x}dx}$

$\mathsf{I~=~\displaystyle\int\limts_{e}^{+\infty}~u^{-2}dx }$

$\mathsf{I~=~-\dfrac{1}{lnx} }$

$\mathsf{I~\Bigg|\limits_{e}^{+\infty}~=~-\dfrac{1}{lnx}\Bigg|\limts_{e}^{+\infty}}~=~-\[\lim_{\rightarrow+\infty} \dfrac{1}{lnx}-\Big(-\dfrac{1}{lne}\Big)\]$

$\mathsf{I~\Bigg|\limts_{e}^{+\infty}~=~-\dfrac{1}{+\infty}+1 }$

$\boxed{\boxed{\mathsf{I~\Bigg|\limts_{e}^{+\infty}=1 }}}}$

Espero ter ajudado bastante!)