(3)Li0(x)=x1−x(4)Li1(x)=−ln(1−x)(5)Li2(x)+Li2(xx−1)=−12ln2(1−x)(6)Li2(x)+Li2(−x)=12Li2(x2)(7)Li2(x)+Li2(1−x)=π26−lnxln(1−x)(8)Lin(x)=∫0xLin−1(x)xdx(9)Lin(x)=(−1)n−1(n