求定积分[0,1]∫xln(x+1)dx原式=[0,1](1/2)∫ln(x+1)dx²=[0,1](1/2){x²ln(x+1)-∫[x²/(x+1)]dx}=[0,1](1/2){x²ln(x+1)-∫[(x-1)+1/(x+1)]dx}=[0,1](1/2){x²ln(x+1)-(x-1)²/2-ln(x+1)}...