二重积分求解，各位高手帮忙，最好详细点解答，谢谢

你好！

积分区域会画吧

∫∫ F(x,y) = ∫[0,2] dx ∫ [x²,2x] x/(y+1) dy
= ∫[0,2] dx * xln(y+1) | [x²,2x]
= ∫[0,2] [ xln(2x+1) - xln(x²+1) ] dx

∫[0,2] xln(2x+1) dx
= 1/2 ∫[0,2] ln(2x+1) dx²
= 1/2 x²ln(2x+1) |[0,2] - 1/2 ∫ x² dln(2x+1)
= 2ln5 - ∫[0,2] x² / (2x+1) dx
= 2ln5 - ∫[0,2] [ (2x-1)/4 + 1/ (4(2x+1)) ] dx
= 2ln5 - [ (x²-x)/4 + 1/8 ln(2x+1) ] [0,2]
= 2ln5 - (1/2 + 1/8 ln5)
= 15/8 ln5 - 1/2

∫[0,2] xln(x²+1) dx
= 1/2 ∫[0,2] ln(x²+1) dx²
=1/2 x²ln(x²+1) |[0,2] - 1/2 ∫[0,2] x² dln(x²+1)
= 2ln5 - ∫[0,2] x³ / (x²+1) dx
= 2ln5 - ∫[0,2] [ x - x/(x²+1) ] dx
= 2ln5 - [ x²/2 - 1/2 ln(x²+1) ] [0,2]
= 2ln5 - (2 - 1/2 ln5)
= 5/2 ln5 - 2

∴原式 = 15/8 ln5 - 1/2 - （5/2 ln5 - 2） = (12 - 5ln5) /8