Question

O valor da integral indefinida abaixo, dx e de;

Answer 1

Boa noite!

Solução!

$\displaystyle \int (\frac{8 x^{4}+9 x^{3} +6 x^{2} -2x+1 }{ x^{2} }-2\sqrt[3]{x}) dx \\\\\\\\ Com a integral montada vamos fazer o MMC.\\\\\\ \displaystyle \int (\frac{8 x^{4}+9 x^{3} +6 x^{2} -2x+1 }{ x^{2} }-2 \sqrt[3]{x}) dx$

Vamos agora simplificar as variavel x!

$\displaystyle \int ( \frac{8 x^{2+1} }{2+1} - \frac{9 x^{1+1} }{1+1} +6x- 2\frac{1}{x}+ \frac{ x^{-2+1} }{ -2+1 }-2 x^{ \frac{1}{3}} )\\\\\\\\\ -2 x^{ \frac{1}{3}}=-2 x^{ \frac{1}{3}+1}$


$\displaystyle \int \frac{8 x^{3} }{3}- \frac{9 x^{2} }{2}+6x-2. \frac{1}{x}+x^{-2}- \frac{2 x^{ \frac{1}{3}+1} }{ \frac{1}{3} +1} \\\\\\\\\ Observac\~ao~~ \frac{1}{x} =ln(x)\\\\\\\ I= \frac{8 x^{3} }{3}- \frac{9 x^{2} }{2}-2. \frac{1}{x}+\frac{ x^{-2+1} }{-2+1} - \frac{2 x^{ \frac{4}{3}} }{ \frac{4}{3}}$

$I= \frac{8 x^{3} }{3}- \frac{9 x^{2} }{2}-2.ln|x|-\frac{{1} }{x} - 2.\dfrac{3}{4}x.^{\dfrac{4}{3}}}}$

$I= \frac{8 x^{3} }{3}- \frac{9 x^{2} }{2}-2.ln|x|- \frac{{1} }{x} - \dfrac{3}{2}x.^{\dfrac{4}{3}}}}$

\boxed{Resposta:C}



Boa noite!
Bons estudos!