Question

Integral da funçao ∫ ((ln x)² + 8)/(x((ln x)³+8)) * dx

Answer 1

É dada a seguinte integral:

$\displaystyle I=\int\dfrac{(\ln x)^2+8}{x((\ln x)^3+8)}\,dx$

Vamos fazer a seguinte substituição:

$u=\ln x\Longrightarrow du=\dfrac{1}{x}dx\Longrightarrow dx=xdu$

Substituindo:

$\displaystyle I=\int\dfrac{u^2+8}{x(u^3+8)}\,xdu\\\\ I=\int\dfrac{u^2+8}{u^3+8}\,du\\\\ I=\underbrace{\int\dfrac{u^2}{u^3+8}\,du}_{I_1}+\underbrace{\int\dfrac{8}{u^3+8}\,du}_{I_2}$

Resolveremos cada uma das integrais destacadas separadamente. Para calcularmos I₁, note que o numerador se assemelha bastante à derivada do denominador. Manipulando a expressão de modo a aproveitarmos esse fato:

$\displaystyle I_1=\int\dfrac{u^2}{u^3+8}\,du\\\\ I_1=\dfrac{1}{3}\int\dfrac{3u^2}{u^3+8}\,du~~~~~~~~\boxed{\dfrac{d}{du}(u^3+8)=3u^2}\\\\ I_1=\dfrac{1}{3}\ln|u^3+8|+C_1$

Agora, desenvolveremos a expressão de I₂:

$\displaystyle I_2=\int\dfrac{8}{u^3+8}\,du\\\\ I_2=8\int\dfrac{1}{u^3+2^3}\,du\\\\ I_2=8\int\dfrac{1}{(u+2)(u^2-2u+4)}\,du$

Quebrando o integrando em frações parciais:

$\dfrac{1}{(u+2)(u^2-2u+4)}=\dfrac{A}{u+2}+\dfrac{Bu+C}{u^2-2u+4}\\\\ \dfrac{1}{(u+2)(u^2-2u+4)}=\dfrac{A(u^2-2u+4)+(Bu+C)(u+2)}{(u+2)(u^2-2u+4)}\\\\ \dfrac{1}{(u+2)(u^2-2u+4)}=\dfrac{(A+B)u^2+(-2A+2B+C)u+(4A+2C)}{(u+2)(u^2-2u+4)}\\\\ \Longrightarrow\begin{cases}A+B=0\to B=-A~~(i)\\-2A+2B+C=0~~(ii)\\4A+2C=1~~~(iii)\end{cases}$

Aplicando i em ii:

$-2A+2B+C=0\\\\ -2A+2(-A)+C=0\\\\ -4A+C=0\\\\ C=4A~~(iv)$

Agora, aplicando iv em iii:

$4A+2C=1\\\\ 4A+2\cdot4A=1\\\\ 4A+8A=12A=1\\\\ \boxed{A=\dfrac{1}{12}}\land\boxed{B=-\dfrac{1}{12}}\land\boxed{C=\dfrac{1}{3}}$

Então:

$\displaystyle I_2=8\left(\int\dfrac{\frac{1}{12}}{u+2}\,du+\int\dfrac{-\frac{1}{12}u+\frac{1}{3}}{u^2-2u+4}\,du\right)\\\\ I_2=\dfrac{2}{3}\int\dfrac{1}{u+2}\,du-\dfrac{1}{3}\int\dfrac{2u-8}{u^2-2u+4}\,du\\\\ I_2=\dfrac{2}{3}\ln|u+2|+C_2-\dfrac{1}{3}\int\dfrac{2u-2}{u^2-2u+4}\,du-\dfrac{1}{3}\int\dfrac{-6}{u^2-2u+4}\,du\\\\ I_2=\dfrac{2}{3}\ln|u+2|+C_2-\dfrac{1}{3}\ln|u^2-2u+4|+C_3+2\underbrace{\int\dfrac{1}{u^2-2u+4}\,du}_{I_3}$

Calculando I₃:

$\displaystyle I_3=\int\dfrac{1}{u^2-2u+4}\,du\\\\ I_3=\int\dfrac{1}{(u^2-2u+1)+3}\,du\\\\ I_3=\int\dfrac{1}{(u-1)^2+3}\,du\\\\ I_3=\dfrac{1}{3}\int\dfrac{1}{(\dfrac{u-1}{\sqrt3})^2+1}\,du\\\\ I_3=\dfrac{\sqrt{3}}{3}\arctan\left(\dfrac{u-1}{\sqrt3}\right)+C_5$

Para simplificar, vamos unir todas as constantes numa só: $C=C_1+C_2+C_3+C_4+C_5$.

Portanto, a resposta final é:

$I=\dfrac{1}{3}\ln|u^3+8|+\dfrac{2}{3}\ln|u+2|-\dfrac{1}{3}\ln|u^2-2u+4|+\dfrac{2\sqrt3}{3}\arctan\left(\dfrac{u-1}{\sqrt3}\right)\\\\ \boxed{I=\dfrac{1}{3}\ln|(\ln x)^3+8|+\dfrac{2}{3}\ln|\ln x+2|-...}\\~\hspace{2.3cm}\boxed{...-\dfrac{1}{3}\ln|(\ln x)^2-2u+4|+\dfrac{2\sqrt{3}}{3}\arctan\left(\dfrac{\ln x-1}{\sqrt3}\right)}$