if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$ 
if $x^5=1$ with $x\neq 1$ then find value of $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$$


So my first observation was x is a non real fifth root of unity. Also $$x^5-1=(x-1)(1+x+x^2+x^3+x^4)=0$$
Thus $$1+x+x^2+x^3+x^4=0$$ I tried using this condition to simplify the above expression but nothing interesting simplified. Please note  i am looking for hints rather than complete solutions.
EDIT:I came to know its a duplicate ,but i feel that the answers given below are different from those in the original.

Note that
$$
\frac1{1+x^n}=\frac12\frac{1+x^{5n}}{1+x^n}=\frac{1-x^n+x^{2n}-x^{3n}+x^{4n}}2\tag1
$$
Applying $(1)$ gives
$$
\begin{align}
\frac{x}{1+x^2}
&=\frac{x-x^3+x^5-x^7+x^9}2\\
&=\frac{x-x^3+1-x^2+x^4}2\tag2
\end{align}
$$
$$
\begin{align}
\frac{x^2}{1+x^4}
&=\frac{x^2-x^6+x^{10}-x^{14}+x^{18}}2\\
&=\frac{x^2-x+1-x^4+x^3}2\tag3
\end{align}
$$
$$
\begin{align}
\frac{x^3}{1+x}
&=\frac{x^3-x^4+x^5-x^6+x^7}2\\
&=\frac{x^3-x^4+1-x+x^2}2\tag4
\end{align}
$$
$$
\begin{align}
\frac{x^4}{1+x^3}
&=\frac{x^4-x^7+x^{10}-x^{13}+x^{16}}2\\
&=\frac{x^4-x^2+1-x^3+x}2\tag5
\end{align}
$$
Each power of $x$ appears twice positive and twice negative, except for $1$ which is always positive. Therefore,
$$
\begin{align}
\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}
&=\frac{1+1+1+1}2\\
&=2\tag6
\end{align}
$$
