Question

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

$\int\limits {(x^2 \sqrt{x^3+1}) } \, dx$

Answer 1

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador:  https://brainly.com.br/tarefa/10635769

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Calcular a integral indefinida:

     $\mathsf{\displaystyle\int\!x^2\cdot \sqrt{x^3+1}\,dx}\\\\\\ \mathsf{=\displaystyle\int\!\frac{1}{3}\cdot 3x^2\cdot \sqrt{x^3+1}\,dx}\\\\\\ \mathsf{=\displaystyle\frac{1}{3}\int\!\sqrt{x^3+1}\cdot 3x^2\,dx}\\\\\\ \mathsf{=\displaystyle\frac{1}{3}\int\!(x^3+1)^{1/2}\cdot 3x^2\,dx}$


Faça a seguinte substituição:

     $\mathsf{x^3+1=u\quad \Rightarrow \quad 3x^2\,dx=du}$


e a integral fica

     $\mathsf{=\displaystyle\frac{1}{3}\int\!u^{1/2}\,du}\\\\\\ \mathsf{=\dfrac{1}{3}\cdot \dfrac{u^{(1/2)+1}}{\frac{1}{2}+1}+C}\\\\\\ \mathsf{=\dfrac{1}{3}\cdot \dfrac{u^{3/2}}{\frac{3}{2}}+C}\\\\\\ \mathsf{=\dfrac{1}{3}\cdot \dfrac{2}{3}\,u^{3/2}+C}\\\\\\ \mathsf{=\dfrac{2}{9}\,u^{3/2}+C}$

     $\mathsf{=\dfrac{2}{9}\cdot (x^3+1)^{3/2}+C}$   <————   esta é a resposta.


Bons estudos! :-)

Answer 2

$\displaystyle\mathsf{\int\,{x}^{2}\sqrt{{x}^{3}+1}dx}=\\\mathsf{\dfrac{1}{3}\int\,3{x}^{2}\sqrt{{x}^{3}+1}dx}</p><p>$

$\mathsf{u={x}^{3}+1\to~du=3{x}^{2}dx}$

Substituindo

$\displaystyle\mathsf{\int\,{x}^{2}\sqrt{{x}^{3}+1}dx}=\\\mathsf{\dfrac{1}{3}\int\,\sqrt{u}du}\\=\mathsf{\dfrac{1}{3}.\dfrac{2}{3}{u}^{\frac{3}{2}}+k}$

Portanto

$\displaystyle\mathsf{\int\,{x}^{2}\sqrt{{x}^{3}+1}dx}=\\\mathsf{\dfrac{2}{9}{({x}^{3}+1)}^{\frac{3}{2}} + k}$