Subiectul al III-lea, 2. b)

∫₀¹ [x²/(f(x) − 6x)] dx = ∫₀¹ [x²/(x³ + 1)] dx = (1/3)∫₀¹ [(x³ + 1)′/(x³ + 1)] dx = [(1/3) ln(x³ + 1)]₀¹ = - 3p = (1/3) ln 2 − (1/3) ln 1 = (1/3) ln 2 - 2p