Question

calcule a integral (lnx)^2 dividido por x

Answer 1

Resposta:

∫ [(ln²x)/x]dx ⇒ Faça u = ln x, de modo que:

• du/dx = 1/x ⇒ du = (1/x)dx.

∫ [(ln²x)/x]dx = ∫ (ln²x . 1/x)dx

∫ [(ln²x)/x]dx = ∫ u²du ⇒ Utilize a regra ∫ xⁿdx = xⁿ⁺¹/(n + 1) + C.

∫ [(ln²x)/x]dx = u²⁺¹/(2 + 1) + C

∫ [(ln²x)/x]dx = u³/3 + C

∫ [(ln²x)/x]dx = (ln³x)/3 + C