$$f(x)=\frac{(x+1)^4}{(35-x^2-2x)^2}, \qquad c=-1$$
I think I solved it:
$$\frac{(x+1)^4}{(35-x^2-2x)^2}=\frac{(x+1)^4}{(-x+5)^2(x+7)^2}$$
$$y=x+1, x=y-1$$
$$\frac{y^4}{(-y+6)^2(y+6)^2}=\frac{y^4}{(36-y^2)^2}=\frac{y^4}{1296}\frac{1}{(1-\frac{y^2}{36})^2}=$$
$$\frac{y^2}{36}\in \langle -1,1 \rangle,y\in \langle -6,6 \rangle$$ $$=\frac{y^4}{36^2}\sum_{n=1}^{\infty}ny^{2n-2}\frac1{36^{n-1}}=\sum_{n=1}^{\infty}ny^{2n+2}\frac1{36^{n+1}}=$$
$$=\sum_{n=1}^{\infty}n(x+1)^{2n+2}\frac1{36^{n+1}}=f(x), x\in \langle -7,5 \rangle $$
Is it correct?