Question 965961
logy = 1/3(log x + log(x+1) + log(x-2) - log(x^2-2) - log(2x+3))

take derivative

y'/y = 1/3 (1/x + 1/x+1 + 1/x-2 - 2x/x^2-2 - 2/2x + 3)

y' = 1/3 (1/x + 1/x+1 + 1/x-2 - 2x/x^2-2 - 2/2x + 3) y

where y= ∛((x(x+1)(x-2))/((x^2-2)(2x+3)))