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
追答∫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