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求不定积分x^2/(x+2)^3dx 用凑微分法应该如何解

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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,
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
再问: 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
再答: ∫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