Making now #y = x-3#
#f(x)=1/(1-x^2) equiv 1/(1-(y+3)^2) = C_1/(1+y+3)+C_2/(1-y-3)#
#=C_1/(y+4)-C_2/(y+2)#
Solving for #C_1,C_2# the condition
#C_1(y+2)-C_2(y+4)=1, forall y in RR# or equivalently
#(C_1-C_2)y+2C_1-4C_2=1, forall y in RR# we obtain
#C_1=C_2 = -1/2# so
#1/(1-x^2) equiv 1/2(1/(y+2)-1/(y+4))=1/2(1/(2(y/2+1))-1/(4(y/4+1)))#
Now considering that for #abs z < 1#
#lim_{n->oo}(z^n-1)/(z-1) = -1/(z-1)=sum_{k = 0}^oo z^n#
restraining #abs(y/2) < 1# then
#1/(y/2+1) equiv -sum_{n=0}^oo (-y/2)^2# and
#1/(y/4+1) equiv -sum_{n=0}^oo (-y/4)^2#
Putting all together
#1/2(1/(y+2)-1/(y+4)) equiv s(y) #
#s(y)= -1/2(1/2 sum_{n=0}^oo (-y/2)^n-1/4sum_{n=0}^oo (-y/4)^n)#
This serie in convergent for #abs y < 2# or equivalently # 1 < x < 5#
Finally, for # 1 < x < 5# we have #f(x) equiv s(x-3)#
#s(x-3) = -1/2sum_{n=0}^oo (-1)^n/2^{n+1}(1-1/2^{n+1})(x-3)^n#
