Question

8) Resolva a integral e diga o método que está usando:

$\int\limits { \frac{x^3-2 \sqrt{x} }{x} } \, dx$

Answer 1

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador:  https://brainly.com.br/tarefa/10635755

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Calcular a integral indefinida:

     $\mathsf{\displaystyle\int\!\frac{x^3-2\sqrt{x}}{x}\,dx}$


Basta simplificar as potências de  x:

     $=\mathsf{\displaystyle\int\!\frac{x^3-2x^{1/2}}{x}\,dx}\\\\\\ =\mathsf{\displaystyle\int\!\left(\frac{x^3}{x}-\frac{2x^{1/2}}{x}\right)dx}\\\\\\ =\mathsf{\displaystyle\int\!\big(x^{3-1}-2x^{(1/2)-1}\big)dx}\\\\\\ =\mathsf{\displaystyle\int\!\big(x^2-2x^{-1/2}\big)dx}\\\\\\ =\mathsf{\displaystyle\int\!x^2\,dx-2\int\!x^{-1/2}\,dx}$


Agora, integre usando a regra da potência:

     $=\mathsf{\dfrac{x^{2+1}}{2+1}-2\cdot \dfrac{x^{(-1/2)+1}}{-\frac{1}{2}+1}+C}\\\\\\ =\mathsf{\dfrac{x^3}{3}-2\cdot \dfrac{x^{1/2}}{\frac{1}{2}}+C}\\\\\\ =\mathsf{\dfrac{x^3}{3}-2\cdot 2x^{1/2}+C}$

     $=\mathsf{\dfrac{x^3}{3}-4\sqrt{x}+C}$   <————   esta é a resposta.


Bons estudos! :-)

Answer 2

$\displaystyle\mathsf{\int\dfrac{x^3-2\sqrt{x}}{x}dx=\int(x^2-2x^{-\frac{1}{2}})dx}$

$\mathsf{\dfrac{1}{3}x^3-4x^{\frac{1}{2}}+k}$