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

$$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3} =$$ $$=\frac{x}{1+x^2}\cdot\frac{x^4}{x^4}+\frac{x^2}{1+x^4}\cdot\frac{x^3}{x^3}+\frac{x^3}{1+x}\cdot \frac{x^2}{x^2}+\frac{x^4}{1+x^3}\cdot\frac{x}{x}= $$ (remember that $x^5=1$, so $x^6=x$ and $x^7=x^2$) $$=\frac{1}{x^4+x}+\frac{1}{x^3+x^2}+\frac{1}{x^2+x^3}+\frac{1}{x+x^4}= 2\left(\frac{1}{x+x^4}+\frac{1}{x^2+x^3}\right)=$$ $$= 2\left(\frac{x^2+x^3+x+x^4}{(x+x^4)(x^2+x^3)} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x^6+x^7} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x+x^2} \right) =2. $$

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, $$ \begin{align} \frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3} &=\frac{1+1+1+1}2\\ &=2\tag6 \end{align} $$