Question

Integral indefinida

$\int\ {} (\frac{(x^2-1)^2}{x^2}) \, dx$

Answer 1

$\displaystyle\int\!\frac{(x^2-1)^2}{x^2}\,dx\\\\\\ =\int\!\frac{(x^2)^2-2\cdot x^2\cdot 1+1^2}{x^2}\,dx\\\\\\ =\int\!\frac{x^4-2x^2+1}{x^2}\,dx\\\\\\ =\int\!\frac{x^4}{x^2}\,dx-\int\!\frac{2x^2}{x^2}\,dx+\int\frac{1}{x^2}\,dx$

$=\displaystyle\int\!x^2\,dx-\int\!2\,dx+\int x^{-2}\,dx\\\\\\ =\frac{x^{2+1}}{2+1}-2x+\frac{x^{-2+1}}{-2+1}+C\\\\\\ =\frac{x^3}{3}-2x+\frac{x^{-1}}{-1}+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{(x^2-1)^2}{x^2}\,dx=\frac{x^3}{3}-2x-\frac{1}{x}+C \end{array}}$


Bons estudos! :-)

Answer 2

$\displaystyle\mathsf{\int\dfrac{(x^2-1)^2}{x^2} \,dx}$

$\displaystyle\mathsf{\int\frac{x^4-2x^2+1}{x^2} \,dx} \\ \mathsf{=\int ({x}^{2} - 2+ \frac{1}{ {x}^{2} })dx} \\ \boxed{\boxed{\mathsf{ \frac{1}{3}{x}^{3} - 2x - \frac{1}{x} + k}}}$