求x（2x-5）的5次幂的不定积分
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x-4 = a(2x-2)+b
x-4 = 2ax - 2a+b
2a = 1 => a = 1/2
b-2a = -4 => b = -3
∫ (x-4)/(x²-2x+5) dx
= (1/2)∫ (2x-2)/(x²-2x+5) dx - 3∫ dx/[(x-1)²+4]
= (1/2)∫ d(x²-2x+5)/(x²-2x+5) - 3∫ d(x-1)/[(x-1)²+2²]
= (1/2)ln|x²-2x+5| - 3*(1/2) arctan[(x-1)/2] + C
= (1/2)ln|x²-2x+5| - (3/2)arctan[(x-1)/2] + C
x-4 = 2ax - 2a+b
2a = 1 => a = 1/2
b-2a = -4 => b = -3
∫ (x-4)/(x²-2x+5) dx
= (1/2)∫ (2x-2)/(x²-2x+5) dx - 3∫ dx/[(x-1)²+4]
= (1/2)∫ d(x²-2x+5)/(x²-2x+5) - 3∫ d(x-1)/[(x-1)²+2²]
= (1/2)ln|x²-2x+5| - 3*(1/2) arctan[(x-1)/2] + C
= (1/2)ln|x²-2x+5| - (3/2)arctan[(x-1)/2] + C
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亲，如果满意，记得采纳哦。
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