Question

[50 pts]

2) Calcular a integral indefinida :

Se puder explicar passo a passo ficaria agradecido.

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

Answer 1

$I=\displaystyle\int\!\left(\dfrac{x^3+2}{x^3} \right )\sqrt{x-\dfrac{1}{x^2}}\,dx\\\\\\ \int\!\left(\dfrac{x^3}{x^3}+\dfrac{2}{x^3} \right )\sqrt{x-\dfrac{1}{x^2}}\,dx\\\\\\ \int\!\left(1+2x^{-3} \right )\sqrt{x-x^{-2}}\,dx\\\\\\ \int\!\sqrt{x-x^{-2}}\cdot \left(1+2x^{-3} \right )dx~~~~~~\mathbf{(i)}$


Fazendo a seguinte substituição:

$x-x^{-2}=u~~\Rightarrow~~(1+2x^{-3})\,dx=du$


Substituindo em $\mathbf{(i)}$ a integral fica

$=\displaystyle\int\!\sqrt{u}\,du\\\\\\ =\int\!u^{1/2}\,du\\\\\\ =\frac{u^{(1/2)+1}}{\frac{1}{2}+1}+C\\\\\\ =\frac{u^{3/2}}{\frac{3}{2}}+C\\\\\\ =\frac{2}{3}\,u^{3/2}+C\\\\\\ =\frac{2}{3}\cdot \left(x-x^{-2} \right )^{3/2}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int\!\left(\dfrac{x^3+2}{x^3} \right )\sqrt{x-\dfrac{1}{x^2}}\,dx=\frac{2}{3}\cdot \left(x-\dfrac{1}{x^2} \right )^{3/2}+C \end{array}}$


Bons estudos! :-)

Answer 2

$\sf \displaystyle \int \left(\frac{x^3+2}{x^3}\right)\sqrt{x-\frac{1}{x^2}\:}dx\\\\\\=\int \sqrt{u}du\\\\\\=\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\\\\\\=\frac{\left(x-\frac{1}{x^2}\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\\\\\\=\frac{2\left(x^3-1\right)^{\frac{3}{2}}}{3x^3}\\\\\\\to \boxed{\sf =\frac{2\left(x^3-1\right)^{\frac{3}{2}}}{3x^3}+C}$