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.
 A: 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,
$$
\begin{align}
\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}
&=\frac{1+1+1+1}2\\
&=2\tag6
\end{align}
$$
A: $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3} =$$
$$=\frac{x}{1+x^2}\cdot\frac{x^4}{x^4}+\frac{x^2}{1+x^4}\cdot\frac{x^3}{x^3}+\frac{x^3}{1+x}\cdot \frac{x^2}{x^2}+\frac{x^4}{1+x^3}\cdot\frac{x}{x}= $$
(remember that $x^5=1$, so $x^6=x$ and $x^7=x^2$)
$$=\frac{1}{x^4+x}+\frac{1}{x^3+x^2}+\frac{1}{x^2+x^3}+\frac{1}{x+x^4}=
2\left(\frac{1}{x+x^4}+\frac{1}{x^2+x^3}\right)=$$
$$= 2\left(\frac{x^2+x^3+x+x^4}{(x+x^4)(x^2+x^3)} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x^6+x^7} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x+x^2} \right) =2.
$$
A: Answer :
((x / (1+x² ))+((x³/ (1+x))=((x+x² +x³ +x⁵) / ((1+x)(1+x²) ))=((x+x² +x³ +x⁵) / (x+x² +x³ +1))=1
Because x^5=1
$((x² / (1+x⁴))+((x⁴/ (1+x³ )) = ((x² +x⁵+x⁴+x⁸) / (1+x³ +x⁴+x^7)) =((x² +1+x⁴+x³) / (1+x³+x⁴+x²)) =1$
Because $x^5 =1$
$\frac{x}{1+x^2 }+\frac{x^3 }{1+x} +\frac{x^2 }{1 +x^4}+\frac{x^4}{1+x^3 }=\frac{x+x^2 +x^3 +x^5}{(1+x)(1+x^2) }+\frac{x^2 +x^5+x^4+x^8}{(1+x^4)(1+x^3) }=\frac{x+x^2 +x^3 +x^5}{x+x^2 +x^3 +1}+\frac{x^2 +1+x^4+x^3 }{1+x^3 +x^4+x^2} =2$
Because $x^5 =1\Rightarrow x^7=x^2$ and $x ^8 = x^3$
