∴(1+x)s'(x) 2.,Q(a-1)2 ax2(a-1)…(a-n+1) =a+ax+ 2,千∵ n =S(x) s'(x ,且s(0)=1. s(x) 1+x 两边积分 xs(x) d a 0a,x∈(-1,1) 0 s(x) 01+x A Ins(x)-Ins(0)=aIn(1+x) 上页(1+ x)s(x)    +   −  − + + +   − =  +  + −1 2 2 2 ! ( 1) ( 1) 2! ( 1) n x n n x x = s(x) , ( ) 1 ( ) s x x s x + =    且 s(0) = 1. 两边积分 , ( ) 1 ( ) 0 0 dx x dx s x x s x x   +  =  x(−1,1) 得 ln s(x) − ln s(0) = ln(1+ x)