Question

A integral ∫▒1/x^2 dx vale

Answer 1

$\mathrm{\int\dfrac{1}{x^2}\ dx=\int x^{-2}\ dx=\dfrac{x^{-2+1}}{-2+1}=\dfrac{x^{-1}}{-1}=\boxed{\mathbf{-\dfrac{1}{x}+C}}}$

Answer 2

Boa tarde Andrea

∫ 1/x² dx = ∫ x^-2 = (-2 + 1)*x^(-2+1) = -1*x^-1 =  -1/x + C