Question

求不定积分x^2/(x+2)^3dx 用凑微分法应该如何解

要求不用有理分式，用第一类换元法（凑微分法）去解题，谢谢了。

Answer 1

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

追问

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

Answer 2

问题不详，无法解答