f(x)=(1/4)*ln[(1+x)-ln(1-x)]+(1/2)*arctanx-x
已知当|x|x] 1/(1+x^2) dx
当|x|x] 1-x^2+x^4-x^6+…… dx= x-x^3/3+x^5/5-x^7/7……
=∑(-1)^n*x^(2n+1)/(2n+1) n from 0 to ∞
所以f(x)=(1/4)*ln[(1+x)/(1-x)]+(1/2)*arctanx-x
=(1/2)∑x^(2n+1) / (2n+1) + (1/2)∑(-1)^n*x^(2n+1)/(2n+1) - x
=[x+x^5/5+x^9/9+x^13/13+……]-x=∑x^(4n+1)/(4n+1) n from 1 to ∞ |x|