Verified answer

∬_R √(3xy+1)dA = ∫₀¹∫₀¹ √(3xy+1) dxdy

= ∫₀¹ [2/3 ⋅ (3xy+1)^(3/2) ⋅ 1/(3y)]₀¹ dy

= (2/9) ∫₀¹ ((3y + 1)^(3/2) - 1) / y dy

Substitute u = √(3y+1), du = 3/(2√(3y+1)) dy, then

= (4/9) ∫₁² (u⁴ - u) / (u² - 1) du

= (4/9) ∫₁² (u³ + u² + u)/(u+1) du (using the fact that (u⁴ - u) = (u-1)(u³ + u² + u) and that (u²-1) = (u+1)(u-1)

= (4/9) ∫₁² u² + 1 - 1/(u+1) du (using long division)

= (4/9) [u³/3 + u - log(u+1)]₁²

= (4/9)(8/3 + 2 - log(3) - 1/3 - 1 + log(2))

= (4/9)(10/3 + log(2/3))