Question

Calcule a integral indefinida

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

Answer 1

Oi Lucas

∫ (x² - 1)²/x²  dx  = ∫ (x⁴ - 2x² + 1)/x²  dx = 

∫ (x² - 2 + 1/x² ) dx = x³/3 - 2x - 1/x + C 

.

Answer 2

$\sf \displaystyle \int \left(\frac{\left(x^2-1\right)^2}{x^2}\right)dx\\\\\\=\frac{\left(x^2-1\right)^2}{x}-\int \frac{2\left(x^2-1\right)\left(x^2+1\right)}{x^2}dx\\\\\\=\frac{\left(x^2-1\right)^2}{x}-2\left(-\frac{\left(x^2-1\right)\left(x^2+1\right)}{x}+\frac{4x^3}{3}\right)\\\\\\\to \boxed{\sf =\frac{\left(x^2-1\right)^2}{x}-2\left(-\frac{\left(x^2-1\right)\left(x^2+1\right)}{x}+\frac{4x^3}{3}\right)+C}$