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This integral requires integration by parts, which follows the formula: ∫ u dv = uv - ∫ v du.We need to choose u and dv such that ∫ v du is simpler to integrate. A common heuristic for choosing 'u' is LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential). In this case, 'ln x' is logarithmic (L) and 'x2' is algebraic (A).Let u = ln x => du = (1/x) dxLet dv = x2 dx => v = ∫ x2 dx = x3/3Now, apply the integration by parts formula:∫ x2 ln x dx = (ln x)(x3/3) - ∫ (x3/3)(1/x) dx∫ x2 ln x dx = (x3/3)ln x - ∫ (x2/3) dxIntegrate the remaining term:∫ (x2/3) dx = (1/3) ∫ x2 dx = (1/3)(x3/3) + C = x3/9 + CSubstitute this back:∫ x2 ln x dx = (x3/3)ln x - x3/9 + CTo match the options, factor out (x3/3):= (x3/3)(ln x - (x3/9)/(x3/3)) + C= (x3/3)(ln x - 1/3) + CThis result exactly matches option C.