Question

∫2/∛x dx como calcular a integral

Answer 1

O resultado dessa integral indefinida é igual a:

   $\Large\displaystyle\text{ \begin{gathered}\sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx=3 \sqrt[3]{\sf x^{2}} +\mathbb{C}\ ,\ \mathbb{C}\in \mathbb{R}\end{gathered} }$

Desejamos calcular a seguinte integral:

    $\Large\displaystyle\text{ \begin{gathered} \sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx\end{gathered} }$

Pela lineariedade, temos que:

$\Large\displaystyle\text{ \begin{gathered} \sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx=2\cdot\int\frac{1}{x^{\frac{1}{3}}}\ dx \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx=2\cdot\int x^{-\frac{1}{3}}\ dx \end{gathered} }$

Agora, vale ressaltar a seguinte propriedade de integração:

     $\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\sf \int x^n\ dx= \frac{x^{n+1}}{n+1} + \mathbb{C}\ ,\ \forall n\neq-1}}\end{gathered} }$

Ficando então:

$\Large\displaystyle\text{ \begin{gathered} \sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx=2\cdot\left[ \frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1} \right]\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx=2\cdot\left[ \frac{x^{\frac{2}{3}}}{\frac{2}{3}} \right]\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx=\diagup\!\!\!{2}\cdot\left[ \frac{3x^{\frac{2}{3}}}{\diagup\!\!\!{2}} \right]\end{gathered} }$

E por fim:

  $\Large\displaystyle\text{ \begin{gathered} \sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx=3x^{\frac{2}{3}}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \therefore \green{\underline{\boxed{\sf \int \frac{2}{\sqrt[3]{\sf x} } \ dx=3 \sqrt[3]{\sf x^{2}} +\mathbb{C}\ ,\ \mathbb{C}\in \mathbb{R}}}}\ \ \ (\checkmark).\end{gathered} }$

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