Question

integral de (x³-8)/ ∛(x^4-32x)^2

Answer 1

$\boxed{\boxed{\int\limits{ \frac{x^3-8}{( \sqrt[3]{x^4-32x} )^2} } \, dx }}$

arrumando o denominador
$( \sqrt[3]{x^4-32x} )^2} =((x^4-32x)^ \frac{1}{3} )^2=(x^4-32x)^ \frac{2}{3}$

então temos
$\boxed{\boxed{ \int\limits { \frac{x^3-8}{(x^4-32x)^ \frac{2}{3} } } \, dx }}$

vou chamar de..
$u= x^4-32x\\\\\\du=(4x^3-32)*dx\\\\\ \frac{du}{(4x^3-32)} =dx$

simplificando isso 
$dx= \frac{du}{4(x^3-8)}$

a expressão fica
${ \frac{(x^3-8)}{u^ \frac{2}{3} } } \, \frac{du}{4(x^3-8)} = \frac{du}{4*u^{ \frac{2}{3}} }$


e a integral
$\int\limits { \frac{1}{u^{ \frac{2}{3} }*4}}.du\\\\\\=\frac{1}{4} \int\limits { \frac{1}{u^{ \frac{2}{3} } }} \, du\\\\\\\boxed{\boxed{{{ \frac{1}{4} \int\limits{u^{- \frac{2}{3} }} \, du }}}}$


agora é só integrar
$\frac{u^{ \frac{-2}{3} +1}}{ \frac{-2}{3}+1 } = \frac{u^{ \frac{1}{3} }}{ \frac{1}{3} }=3u^{ \frac{1}{3} }= \sqrt[3]{u}$

multiplicando isso pelo 1/4 que estava do lado d fora da integral
$\frac{1}{4}* 3 \sqrt[3]{u} = \frac{3 \sqrt[3]{u} }{4}$

u= x^4-32x

então 
$\boxed{\boxed{{ \frac{3 \sqrt[3]{x^4-32x} }{4} }}}$

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
RESPOSTA 
$\boxed{\boxed{\int\limits{ \frac{x^3-8}{( \sqrt[3]{x^4-32x} )^2} } \, dx = \frac{3 \sqrt[3]{x^4-32x} }{4}+K }}}$

K = constante