Respostas

2014-08-16T04:05:57-03:00
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) }}
1 5 1
ah sim, eu tava tentando udu, por isso n estava conseguindo nao sabia o q substituir hahaha vlw