求函数ln(x+√(1+x^2))在x=0处的幂级数展式,并求展开式成立的区间

(ln(x+√(1+x^2)))'=1/(√(1+x^2))=(1+x^2)^(-1/2)(1+x^2)^(-1/2)=1-(1/2)x^2+(-1/2)(-1/2-1)/2!(x^4)+(-1/2)(-1/2-1)(-1/2-2)/3!(x^6)+...=1-(1/2)x^2+(-1/2)(-3/2)/2!(x^4)+(-1/2)(-3/2)(-5/2)/3!(x^6)+...=1-(1/2...