jmc

When $x+x^3+x^9+x^{27}+x^{81}+x^{243}$ is divided by $x^2-1,$ the remainder is of the form $ax+b,$ so $$x+x^3+x^9+x^{27}+x^{81}+x^{243}=(x^2-1)q(x)+ax+b$$for some polynomial $q(x).$

Setting $x=1,$ we get $$6=a+b.$$Setting $x=-1,$ we get $$-6=-a+b.$$Solving this system, we find $a=6$ and $b=0,$ so the remainder is $\boxed{6x}.$