To evaluate the definite integral ∫2(2x-3)2/3 dx, we can use the following formula: ∫un du = (un+1)/(n+1) + C where C is the constant of integration. Using this formula, we can rewrite the integral as: ∫2(2x-3)2/3 dx = 2 ∫(2x-3)2/3 dx Let u = 2x - 3, then du/dx = 2 and dx = du/2. Substituting this into the integral, we get: 2 ∫(2x-3)2/3 dx = 2 ∫u2/3 (du/2) = ∫u2/3 du Using the formula, we get: ∫u2/3 du = (u5/3)/(5/3) + C Substituting back for u, we get: ∫2(2x-3)2/3 dx = (2(2x-3)5/3)/(5/3) + C Simplifying, we get: ∫2(2x-3)2/3 dx = (6(2x-3)5/3)/5 + C Therefore, the correct answer is option A: 3/5(2x-3)5/3 + k.