求不定积分详细解答下 呼呼~

_yueyaer 2023-03-17 11:33

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  _B吇

  2023-03-17 11:35

  原式=1/2∫ln(x+1)dx²=1/2x²ln(x+1)-1/2∫x²dln(x+1)=[x²ln(x+1)]/2-1/2∫x²/(x+1)dx单独求∫x²/(x+1)dx=∫(x²-1+1)/(x+1)dx=∫[(x²-1)/(x+1)+1/(x+1)]dx=∫[(x-1)+1/(x+1)]dx=∫(x-1)dx+∫dx/(x+1)=x²/2-x+ln(x+1)+C所以原式=[x²ln(x+1)]/2-x²/4+x/2-1/2*ln(x+1)+C =[(x²-1)/2]ln(x+1)-x²/4+x/2+C

  ∫xln（x+1）dx＝1/2x²(lnx－1/2)＋C
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