微积分数学题 ！急！追加分！∫(x^2-3x+2)/(x+1)dx

解:原式=∫(x^2+x-4x-4+6)/(x+1)dx
=∫x-4+[6/(x+1)]dx
=1/2x^2-4x+6lg(x+1)+c

∫(x^2-3x+2)/(x+1)dx
=∫(x^2+x-4x+2)/(x+1)dx
=∫[x-(4x-2)/(x+1)]dx
=∫[x-(4x+4-6)/(x+1)]dx
=∫[x-4+6/(x+1)]dx
=(1/2)x^2-4x+6ln|x+1|+C

答案是X^2/2-4X+6ln(X+1)+C

关键在于拆项：
原式=∫[(x+1)^2-5(x+1)+6]/(x+1)dx
=∫(x+1)dx-∫5dx+6∫1/(x+1)dx
=1/2x^2+x-5x+6ln(x+1)+c
=1/2x^2-4x+6ln(x+1)+c