Let y=x1+x2
then f=y+y33+y55+...
=12log(1+y1−y)
=12log⎛⎜
⎜⎝1+x1+x21−x1+x2⎞⎟
⎟⎠
=12log(1+x+x21−x+x2) ...(1)
and g=x−23x3+15x5+17x7+29x9+111x11+113x13+215x15+...
=x−(1−13)x3+15x5+17x7−13(1−13)x9+111x11+113x13−15(1−13)x15+...
=(x+x33+x55+x77+...)−(x3+x93+x155+...∞)
=12log(1+x1−x)−12log(1+x31−x3)
=12log[(1+x)(1−x).(1+x3)(1−x3)]
=12log[(1+x)(1−x)(1+x+x2)(1−x)(1+x)(1−x+x2)]
=12log(1+x+x21−x+x2)
from (1) and (2), we get
f≡g
⇒c=3