1,原式＝ S（x^2-3x+9-27/(x+3) ) dx
= 1/3*x^3- 3/2* x^2+ 9x-27*Ln(x+3) + C
2,原式＝ S [3(x+1)-4]/[(x+1)^2 + 4] dx
= S 3(x+1)/[(x+1)^2 + 4]dx - S 4/[(x+1)^2 + 4]dx
= 3/2 S d[(x+1)/2] ^2 /{ [(x+1)/2]^2 +1}
+ 2 S d[(x+1)/2]/{ [(x+1)/2]^2 +1}
= 3/2 Ln[(x+1)^2/4 +1] - 2arctg[(x+1)/2] + C
3,令x+1=t^3
原式＝ 3 S t^2/(t+1) dt = 3 S (t-1)dt + 3 S 1/(t+1)dt
= 3/2 t^2 -3t + 3 Ln(t+1) +C
= 3/2 (x+1)^2/3 - 3 (x+1)^1/3
+ 3 Ln((x+1)^1/3 +1) + C
符号有点混乱,不过只要思路看清楚就好了.