(1+x)s(x) 2x+x2+…xc2(a-1)·(a-n+xn1+ a+aX 2! as(x) s(r ,且s(0)=1. s(x)1+x 两边积分 xs(x) x a dx,x∈(-1,1) 0 s(r) 01+x L Ins(x)-In s(0)=aIn(1+x),(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)