定积分,求简便算法
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  • ∫ (x-1)^4/(x+1)^4 dx

    =∫ (x+1-2)^4/(x+1)^4 dx 分子用二项式展开

    =∫ [(x+1)^4-8(x+1)³+24(x+1)²-32(x+1)+16)/(x+1)^4 dx

    =∫ 1 dx - 8∫ 1/(x+1) dx + 24∫ 1/(x+1)² dx - 32∫ 1/(x+1)³ dx + 16∫ 1/(x+1)^4 dx

    =x - 8ln(x+1) - 24/(x+1) + 16/(x+1)² - (16/3)(1/(x+1)³) |[0→1]

    =1 - 8ln2 - 12 + 4 - 2/3 - 0 + 0 + 24 - 16 + 16/3

    =17/3 - 8ln2

    已用数学软件验算过,结果正确.

    你发的同样的另一个题,我也答了,被百度吞了,郁闷中.

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