Question

Alguém pode responder com detalhes, Integral [x.(raiz cúbica x) + 1]^2

Answer 1

Calcular a integral indefinida:

     $\displaystyle\int (x\,^3\!\!\!\!\sqrt{x}+1)^2\,dx\\\\\\ =\int \big(\,^3\!\!\!\sqrt{x^3}\cdot \,^3\!\!\!\!\sqrt{x}+1\big)^2\,dx\\\\\\ =\int \big(\,^3\!\!\!\sqrt{x^3\cdot x}+1\big)^2\,dx\\\\\\ =\int \big(\,^3\!\!\!\sqrt{x^4}+1\big)^2\,dx\\\\\\ =\int \big(x^{4/3}+1\big)^2\,dx$


Agora expanda o quadrado da soma, usando produtos notáveis:

     •  (a + b)² = a² + 2ab + b²


e a integral fica

     $\displaystyle=\int \big[(x^{4/3})^2+2\cdot x^{4/3}\cdot 1+1^2\big]dx\\\\\\ =\int \big[x^{8/3}+2x^{4/3}+1\big]dx\\\\\\ =\int x^{8/3}\,dx+2\int x^{4/3}\,dx+\int 1\,dx$


Agora aplique a regra para integrar potências:

     •  $\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\qquad\quad \mathsf{para~}n\ne -1$


e a integral fica

     $=\dfrac{x^{(8/3)+1}}{\frac{8}{3}+1}+2\cdot \dfrac{x^{(4/3)+1}}{\frac{4}{3}+1}+x+C\\\\\\ =\dfrac{x^{11/3}}{\frac{11}{3}}+2\cdot \dfrac{x^{7/3}}{\frac{7}{3}}+x+C\\\\\\ =\dfrac{3}{11}\,x^{11/3}+2\cdot \dfrac{3}{7}\,x^{7/3}+x+C$

     $=\dfrac{3}{11}\,x^{11/3}+\dfrac{6}{7}\,x^{7/3}+x+C\quad\longleftarrow\quad \mathsf{resposta.}$


Dúvidas? Comente.


Bons estudos! :-)