这个怎么得不定积分啊？
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令 t = √(2x+3), 则 x = (t^2-3)/2, dx = tdt, 得
I = ∫ x^4 √(2x+3) dx = (1/16)∫(t^2-3)^4 t^2 dt
= (1/16)∫(t^10 - 12t^8 + 54t^6 - 108t^4 + 81t^2) dt
= (1/16)[(1/11)t^11 - (4/3)t^9 + (54/7)t^7 - (108/5)t^5 + 27t^3] + C
= (1/16)t^3[(1/11)t^8 - (4/3)t^6 + (54/7)t^4 - (108/5)t^2 + 27] + C
= (1/16)(2x+3)^(3/2)[(1/11)(2x+3)^4 - (4/3)(2x+3)^3
+ (54/7)(2x+3)^2 - (108/5)(2x+3) + 27] + C
I = ∫ x^4 √(2x+3) dx = (1/16)∫(t^2-3)^4 t^2 dt
= (1/16)∫(t^10 - 12t^8 + 54t^6 - 108t^4 + 81t^2) dt
= (1/16)[(1/11)t^11 - (4/3)t^9 + (54/7)t^7 - (108/5)t^5 + 27t^3] + C
= (1/16)t^3[(1/11)t^8 - (4/3)t^6 + (54/7)t^4 - (108/5)t^2 + 27] + C
= (1/16)(2x+3)^(3/2)[(1/11)(2x+3)^4 - (4/3)(2x+3)^3
+ (54/7)(2x+3)^2 - (108/5)(2x+3) + 27] + C
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