Is the answer still 2?

Method 1:

$\begin{aligned} S & = \frac x{1+x^2} + \frac {x^2}{1+x^4} + \frac {x^3}{1+x} + \frac {x^4}{1+x^3} \\ & = \frac x{1+x^2} + \frac {{\color{#3D99F6}x} \cdot x^2}{{\color{#3D99F6}x} \cdot +{\color{#3D99F6}x} \cdot x^4} + \frac {x^3}{1+x} + \frac {{\color{#3D99F6}x^2} \cdot x^4}{{\color{#3D99F6}x^2} \cdot +{\color{#3D99F6}x^2} \cdot x^3} & \small \color{#3D99F6} \text{Note that } x^5 = 1 \\ & = \frac x{1+x^2} + \frac {\color{#3D99F6}x^3}{\color{#3D99F6}1+x} + \frac {x^3}{1+x} + \frac {\color{#3D99F6}x}{\color{#3D99F6}1 + x^2} \\ & = 2 \left(\frac x{1+x^2} + \frac {x^3}{1+x} \right) \\ & = 2 \left(\frac {x+x^2+x^3+{\color{#3D99F6}x^5}}{(1+x^2)(1+x)} \right) \\ & = 2 \left(\frac {{\color{#3D99F6}1}+x+x^2+x^3 }{1+x + x^2+x^3} \right) \\ & = \boxed{2} \end{aligned}$

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Method 2:

$x^5=1$ , $x \ne 1$ means that $x$ is the fifth complex roots of unity or $x = e^{\frac {2k\pi}5 i}$ , where $k = 1, 2, 3, 4$ . Then,

$\begin{aligned} S & = \frac x{1+x^2} + \frac {x^2}{1+x^4} + \frac {x^3}{1+{\color{#3D99F6}x}} + \frac {x^4}{1+{\color{#3D99F6}x^3}} & \small \color{#3D99F6} \text{Note that }x^5 = 1 \\ & = \frac x{1+x^2} + \frac {x^2}{1+x^4} + \frac {x^3}{1+{\color{#3D99F6}x^6}} + \frac {x^4}{1+{\color{#3D99F6}x^8}} \\ & = \frac 1{x+x^{-1}} + \frac 1{x^2+x^{-2}} + \frac 1{x^3+x^{-3}} + \frac 1{x^4+x^{-4}} \\ & = \frac 1{e^{\frac {2\pi}5 i}+e^{-\frac {2\pi}5 i}} + \frac 1{e^{\frac {4\pi}5 i}+e^{-\frac {4\pi}5 i}} + \frac 1{e^{\frac {6\pi}5 i}+e^{-\frac {6\pi}5 i}} + \frac 1{e^{\frac {8\pi}5 i}+e^{-\frac {8\pi}5 i}} \\ & = \frac 12 \left(\frac 1{\cos \frac {2\pi}5} + \frac 1{\cos \frac {4\pi}5} +\frac 1{ \cos \frac {6\pi}5} + \frac 1{\cos \frac {8\pi}5} \right) \\ & = \frac 12 \left(\frac 1{\cos \frac {2\pi}5} - \frac 1{\cos \frac {\pi}5} - \frac 1{\cos \frac {\pi}5} + \frac 1{\cos \frac {2\pi}5} \right) \\ & = \frac 1{\cos \frac {2\pi}5} - \frac 1{\cos \frac {\pi}5} = \frac {\cos \frac {\pi}5 - \cos \frac {2\pi}5}{\cos \frac {\pi}5 \cos \frac {2\pi}5} = \frac {- \cos \frac {4\pi}5 - \cos \frac {2\pi}5}{\cos \frac {\pi}5 \cos \frac {2\pi}5} \\ & = \frac {\frac 12}{\cos \frac {\pi}5 \cos \frac {2\pi}5} = \frac {\color{#3D99F6}\sin \frac \pi 5}{2 {\color{#3D99F6}\sin \frac \pi 5} \cos \frac {\pi}5 \cos \frac {2\pi}5} \\ & = \frac {\sin \frac \pi 5}{\sin \frac {2\pi} 5 \cos \frac {2\pi}5} = \frac {2 \sin \frac \pi 5}{\sin \frac {4\pi} 5} = \frac {2 \sin \frac \pi 5}{\sin \frac {\pi} 5} = \boxed{2} \end{aligned}$

Enough with the complex algebra. There is a fast way to solve this.

Multiplying $\dfrac {x}{1+{x}^{2}} \cdot \dfrac {x^3}{x^3} +\dfrac {x^2}{1+{x}^{4}} \cdot \dfrac {x}{x} +\dfrac {x^3}{1+{x}} +\dfrac {x^4}{1+x^3}$

Which becomes $\dfrac { x^4 }{ x^3+{ x }^{ 5 } } +\dfrac { x^{ 3 } }{ x+{ x }^{ 5 } }+\dfrac { x^{ 3 } }{ 1+{ x } } +\dfrac { x^{ 4 } }{ 1+x^{ 3 } }$ , but $x^5=1$ .

Hence $\dfrac { x^4 }{ 1+x^3 } +\dfrac { x^{ 3 } }{ 1+x }+\dfrac { x^{ 3 } }{ 1+{ x } } +\dfrac { x^{ 4 } }{ 1+x^{ 3 } } \Rightarrow 2 \cdot \left( \dfrac { x^{ 3 } }{ 1+{ x } } +\dfrac { x^{ 4 } }{ 1+x^{ 3 } } \right)$ .

Now you can try to sum the fractions, calculate the LCM, but this goes against my train of thought, which is multiply one of the fractions to be able to sum then up.

Multiply the first fraction $2 \cdot \left( \dfrac { x^{ 3 } }{ 1+ x } \cdot \dfrac {x^2-x+1}{x^2-x+1} +\dfrac { x^{ 4 } }{ 1+x^{ 3 } } \right)$

By the identity $x^3+1=(x+1)(x^2-x+1)$ ,

$\begin{aligned} 2 \cdot \left( \dfrac { x^{ 3 } \cdot x^2-x+1 }{ 1+x^3} +\dfrac { x^{ 4 } }{ 1+x^{ 3 } } \right) \\ \\ 2 \cdot \left( \dfrac { x^5 - x^4 +x^3 }{ 1+x^3} +\dfrac { x^{ 4 } }{ 1+x^{ 3 } } \right) \\ \\ 2 \cdot \left( \dfrac { x^5 - x^4 +x^3 + x^4 }{ 1+x^3} \right) \\ \\ 2 \cdot \left( \dfrac { 1 +x^3 }{ 1+x^3} \right) \\ 2 \cdot \left( \frac {1}{1}\right) = \boxed { 2 } \end{aligned}$

Now you guys could be wondering, how could I have this visualization? Algebraic intuition, and such. It will be polished only by training and time. Now some of my motivations to proceed this way: The apparent symmetry of the fractions, the order of powers of x and finally my approach, that you saw I didn't change, even after the first multiplications, i.e. if you start solving a problem by Approach A, you should keep this approach, which is more likely to solve the whole problem.