Question

Pessoas nao estou conseguindo fazer a integral por udu dessa conta.. Isso è td que consegui, alguem consegue me dar o proximo passo ? Onde substituo?

Answer 1

primeiro vamos arrumar a função
$f(x)= \frac{8x-5}{ \sqrt[3]{x} }\\\\ = \frac{8x-5}{ x^\frac{1}{3} }\\\\= \frac{8x}{x^\frac{1}{3}} + \frac{5}{x^\frac{1}{3}}\\\\= {8x^{ 1-\frac{1}{3} }}+ \frac{5}{x^\frac{1}{3}}\\\\\\\ \boxed{\boxed{ f(x)=8x^ \frac{2}{3} + \frac{5}{x^\frac{1}{3}}}}$
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agora escrevendo a integral 
$\int\limits{8x^ \frac{2}{3} + \frac{5}{x^\frac{1}{3}}} \, dx =\boxed{\boxed{ \int\limits {8^{ \frac{2}{3} }} \, dx + \int\limits { \frac{5}{x^\frac{1}{3}} } \, dx }}$

como 8..e o 5 são constantes..posso coloca-los do lado d fora da integral
fica
$\boxed{\boxed{(8\int\limits {x^{ \frac{2}{3} }} \, dx )+(5 \int\limits { \frac{1}{x^\frac{1}{3}} } \, dx )}}}}$

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resolvendo a primeira integral
$\frac{x^{\frac{2}{3} +1}}{\frac{2}{3} +1} = \frac{x^{ \frac{5}{3} }}{ \frac{5}{3} } =\boxed{ \frac{3x^{ \frac{5}{3} }}{5} }$
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resolvendo a segunda integral
$\frac{1}{x^{ \frac{1}{3} }} =x^{- \frac{1}{3} }\to \frac{x^{- \frac{1}{3} +1}}{- \frac{1}{3} +1}} = \frac{x^{ \frac{2}{3} }}{ \frac{2}{3} } =\boxed{ \frac{3x^{ \frac{2}{3} }}{2} }$
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agora temos

$8*( \frac{3x^{ \frac{5}{3} }}{5}) + 5*( \frac{3x^{ \frac{2}{3} }}{2})\\\\=( \frac{24x^{ \frac{5}{3} }}{5}) + ( \frac{15x^{ \frac{2}{3} }}{2})\\\\=( \frac{2*24x^{ \frac{5}{3} }}{5*2}) + ( \frac{15x^{ \frac{2}{3} }}{2*5})\\\\=\boxed{ \frac{48x^{ \frac{5}{3} }+75x^{ \frac{2}{3} }}{10} } }$

podemos colocar $3x^{ \frac{2}{3} }$ em evidencia no numerador

$\frac{ 3x^ \frac{2}{3}*(16x- 25) }{10} \\\\\\\ \boxed{\boxed{ \frac{3 \sqrt[3]{x^2} }{10}*(16x-25) }}$