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x³/(x - 1)^10
= x²[(x - 1) + 1]/(x - 1)^10
= x²/慧空(x - 1)^9 + x²/(x - 1)^10
= x[(x - 1) + 1]/(x - 1)^9 + x[(x - 1) + 1]/(x - 1)^10
= x/(x - 1)^8 + 2x/(x - 1)^9 + x/(x - 1)^10
= [(x - 1) + 1]/(x - 1)^8 +2[(x - 1) + 1]/(x - 1)^9 + [(x - 1) + 1]/(x - 1)^10
= 1/(x - 1)^7 + 1/(x - 1)^8 + 2/(x - 1)^8 + 2/(x - 1)^9 + 1/(x - 1)^9 + 1/(x - 1)^10
= 1/(x - 1)^7 + 3/(x - 1)^8 + 3/(x - 1)^9 + 1/(x - 1)^10
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∫ x³/(x - 1)^10 dx
= ∫ [1/(x - 1)^7 + 3/(x - 1)^8 + 3/(x - 1)^9 + 1/(x - 1)^10] d(x - 1)
= -1/[6(x - 1)^6] - 3/[7(x - 1)^7] - 3/败让[8(x - 1)^8] - 1/[9(x - 1)^9] + C
= (-84x³察碧局 + 36x² - 9x + 1)/[504(x - 1)^9] + C
= x²[(x - 1) + 1]/(x - 1)^10
= x²/慧空(x - 1)^9 + x²/(x - 1)^10
= x[(x - 1) + 1]/(x - 1)^9 + x[(x - 1) + 1]/(x - 1)^10
= x/(x - 1)^8 + 2x/(x - 1)^9 + x/(x - 1)^10
= [(x - 1) + 1]/(x - 1)^8 +2[(x - 1) + 1]/(x - 1)^9 + [(x - 1) + 1]/(x - 1)^10
= 1/(x - 1)^7 + 1/(x - 1)^8 + 2/(x - 1)^8 + 2/(x - 1)^9 + 1/(x - 1)^9 + 1/(x - 1)^10
= 1/(x - 1)^7 + 3/(x - 1)^8 + 3/(x - 1)^9 + 1/(x - 1)^10
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∫ x³/(x - 1)^10 dx
= ∫ [1/(x - 1)^7 + 3/(x - 1)^8 + 3/(x - 1)^9 + 1/(x - 1)^10] d(x - 1)
= -1/[6(x - 1)^6] - 3/[7(x - 1)^7] - 3/败让[8(x - 1)^8] - 1/[9(x - 1)^9] + C
= (-84x³察碧局 + 36x² - 9x + 1)/[504(x - 1)^9] + C
追问
能告诉我你的思路么?
追答
在分子中弄个x - 1,使分母降次之余也会简化分子中的x项
这种方法是部分分式
也可以令
x³/(x - 1)^10 = A/(x - 1)^10 + B/(x - 1)^9 + C/(x - 1)^8 + D/(x - 1)^7
分母是10次方,分子是3次方,相差7次方,所以将分母降三个次方就可。
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