<s>[INST] Solve this equation $\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}$ 
Solve this equation
  $$\frac{1}{x+\sqrt{x^2-1}}=\frac{1}{4x}+\frac{3x}{2x^2+2}$$

This equation has solutions $x=\pm 1$ but I cannot find a solution.
Do you think using an inequality will is the way forward?
 [/INST] $$\frac{1}{x+\sqrt{x^2-1}}=x-\sqrt{x^2-1}=\frac{7x^2+1}{4x^3+4x}$$
$$\sqrt{x^2-1}=\frac{4x^4-3x^2-1}{4x^3+4x}=\frac{(4x^2+1)\cdot(x-1)\cdot(x+1)}{4x\cdot(x^2+1)}$$
$$\sqrt{x^2-1}\geq0,\space so\space\space x \in [-1,0] \cup \{1\}$$
$$x^2-1=(x-1)\cdot(x+1)=\frac{(4x^2+1)^2\cdot(x-1)^2\cdot(x+1)^2}{16x^2\cdot(x^2+1)^2}$$
$$(x-1)\cdot(x+1)\cdot(\frac{(4x^2+1)^2\cdot(x-1)\cdot(x+1)}{16x^2\cdot(x^2+1)^2}-1)=0$$
$$\rightarrow x \in \{-1,1\}$$
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