Question

Resolver:
$\boxed{\rm \huge \int ^{ \infty } _{1} \cfrac{dx}{x + {x}^{a} } }$
O assunto é integrais impróprios ;D
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Answer 1

✅ Se $\largea>1$ , a integral dada converge para  $\large\begin{gathered} \frac{\ln2}{\alpha-1}\end{gathered}$ . Caso a $\largea\leqslant1$, a integral diverge.

❏ Para resolver essa integral , iremos utilizar o método da substituição simples, mas para isso, vamos dividir a função f(x) por x^a, ficando então mais fácil de se resolver por tal método.

❏ Tendo isso em mente, vamos fazer primeiramente essa integral adotando $\large\begin{gathered} \alpha \neq 1\end{gathered}$, ou seja, faremos três soluções, sendo elas para $\large\begin{gathered} \alpha > 1\end{gathered}$, $\large\begin{gathered} \alpha < 1\end{gathered}$ e $\large\begin{gathered} a= 1\end{gathered}$.

❏ Primeiramente, iremos resolver para $\large\begin{gathered} \alpha \neq 1\end{gathered}$, ficando então:

$\Large\displaystyle\text{ \begin{gathered} F(x)=\int _1^{\infty}\frac{1}{x+x^a}dx = \int _1^{\infty}\frac{\left(\frac{1}{x^a}\right)}{\left(\frac{x}{x^a} +\frac{x^a}{x^a }\right) }dx \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} F(x)= \int _1^{\infty}\frac{\left(\frac{1}{x^a}\right)}{\left(\frac{x}{x^a} +\frac{x^a}{x^a }\right) } dx =\int_1^{\infty}\frac{x^{-a}}{x^{1-a}+1}dx \end{gathered} }$

❏ Aplicando o método da substituição, temos que:

$\Large\displaystyle\begin{gathered}u= x^{1-a}+1\implies du= (1-a)\cdot x^{\!\diagup\!\!\!\!1-a-\!\diagup\!\!\!\!1}dx \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered} \frac{du}{(1-a)\cdot \!\diagup\!\!\!\!x^{-a}}\cdot \!\diagup\!\!\!\!x^{-a}=x^{-a}dx \end{gathered} }$

$\Large\displaystyle\begin{gathered} \frac{du}{(1-a)}=x^{-a}dx \end{gathered}$

• (✍️)  Substituindo, temos:

 $\Large\displaystyle\begin{gathered} F(x)= \int_1^{\infty}\frac{1}{u}\cdot \frac{du}{(1-a)} \end{gathered}$

 $\Large\displaystyle\begin{gathered} F(x)= \frac{1}{(1-a)}\cdot \int_1^{\infty}\frac{du}{u} \end{gathered}$

 $\Large\displaystyle\begin{gathered} F(x)= \frac{1}{(1-a)}\cdot \left( \ln|u| \right)\bigg|_1^{\infty}\end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= \frac{1}{(1-a)}\cdot \left( \ln|x^{1-a}+1| \right)\bigg|_1^{\infty}\end{gathered}$

❏ Agora, vamos analisar esse TFC ( Teorema Fundamental do Cálculo ). Adotando a > 1 , temos que:

$\Large\displaystyle\begin{gathered} F(x)= \frac{1}{(1-a)}\cdot \left( \ln\left|\frac{1}{\infty}+1\right|-\ln \left| \frac{1}{1}+1\right| \right)\end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= \frac{1}{(1-a)}\cdot \left( \ln\left|1\right|-\ln \left| 2\right| \right)\end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= -\frac{\ln |2|}{(1-a)} \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\therefore F(x)= \frac{ \ln|x^{1-a}+1| }{(1-a)}\bigg|_1^{\infty}\ \forall a>1 \ =\ \green{\boxed{\frac{ \ln 2 }{a-1}}}\ \checkmark\end{gathered} }$

❏ Vamos agora fazer analizar esse TFC adotando a < 1, ficando então:

$\Large\displaystyle\begin{gathered} F(x)= \frac{1}{(1-a)}\cdot \left(\ln\left| \infty+1\right|-\ln \left| 1+1\right| \right)\end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= \frac{1}{(1-a)}\cdot \infty-\ln \left| 2\right| \end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= \frac{1}{(1-a)}\cdot \infty\end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= \infty\end{gathered}$      

     $\Large\displaystyle\text{ \begin{gathered} \therefore F(x)=\frac{ \ln|x^{1-a}+1| }{(1-a)}\bigg|_1^{\infty}\ \forall a<1 \ =\ \green{\boxed{\infty}}\ \checkmark\end{gathered} }$

❏ E por fim, adotando a = 1, temos que:

$\Large\displaystyle\begin{gathered} F(x)= \int^{\infty}_{1}\frac{1}{x+x}dx \end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= \int^{\infty}_{1}\frac{1}{2x}dx \end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= \frac{1}{2}\cdot \int^{\infty}_{1}\frac{dx}{x} \end{gathered}$

$\Large\displaystyle\begin{gathered} F(x)= \frac{1}{2}\cdot \ln |\infty|-\ln|1|\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered} \therefore F(x)=\green{\boxed{\infty }}\ \checkmark\end{gathered} }$

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