Derivación: (xs (x))' = 2 (1+ x 2 +x 4+...)= 2/(1-x ^ 2) (aquí está x ^ 2, no x).
xs(x)=∫2dx/(1-x^2)=ln|(x+1)/(x-1)|+c,c=0
s(x)= ln |(x+1)/(x-1)|/x
Derivación: (xs (x))' = 2 (1+ x 2 +x 4+...)= 2/(1-x ^ 2) (aquí está x ^ 2, no x).
xs(x)=∫2dx/(1-x^2)=ln|(x+1)/(x-1)|+c,c=0
s(x)= ln |(x+1)/(x-1)|/x