积分1/(t+1)^2(t-2)dt

2个回答

  • 设1/[(t+1)²(t-2)]=A/(t+1) + B/(t+1)² + C/(t-2)

    右边合并后与左边比较系数得:A=-1/9,B=-1/3,C=1/9

    ∫ 1/[(t+1)²(t-2)] dt

    =-(1/9)∫ 1/(t+1) dt - (1/3)∫ 1/(t+1)² dt + (1/9)∫ 1/(t-2) dt

    =-(1/9)ln|t+1| + (1/3)/(t+1) + (1/9)ln|t-2| + C

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