令x²/(x+2)³=A/(x+2)+B/(x+2)²+C/(x+2)³解得A=1,B=-4,C=4原式=∫dx/(x+2) - 4∫dx/(x+2)² + 4∫dx/(x+2)³=ln|x+2| + 4/(x+2) - 2/(x+2)² + C=(4x+6)/(x+2)² + ln|x+2| + C...首先先谢谢你了，不过 用第一类换元法（凑微分法）应该如何去解题∫x²/(x+2)³ dx=∫x² d[-1/2(x+2)²]=(-1/2)∫x² d[1/(x+2)²]=(-1/2)*x²/(x+2)² + (1/2)∫1/(x+2)² d(x²)，这里运用分部积分法=(-1/2)*x²/(x+2)² + ∫x/(x+2)² dx=(-1/2)*x²/(x+2)² + ∫(x+2-2)/(x+2)² dx=(-1/2)*x²/(x+2)² + ∫[1/(x+2)-2/(x+2)²] dx=(-1/2)*x²/(x+2)² + ∫d(x+2)/(x+2) - 2∫d(x+2)/(x+2)²=(-1/2)*x²/(x+2)² + ln|x+2| - [-2/(x+2)] + C=-x²/[2(x+2)²] + ln|x+2| + 2/(x+2) + C