14. Solve the inequality
\[
\begin{array}{l}
\log _{2}\left(x^{12}+3 x^{10}+5 x^{8}+3 x^{6}+1\right) \\
<1+\log _{2}\left(x^{4}+1\right) .
\end{array}
\]
14. Solution 1: From
$$
1+\log _{2}\left(x^{4}+1\right)=\log _{2}\left(2 x^{4}+2\right),
$$

and $\log _{2} y$ is an increasing function on $(0,+\infty)$, the original inequality is equivalent to
$$
x^{12}+3 x^{10}+5 x^{8}+3 x^{6}+1x^{6}+3 x^{4}+3 x^{2}+1+2 x^{2}+2 \\
=\left(x^{2}+1\right)^{3}+2\left(x^{2}+1\right), \\
\left(\frac{1}{x^{2}}\right)^{3}+2\left(\frac{1}{x^{2}}\right)>\left(x^{2}+1\right)^{3}+2\left(x^{2}+1\right) . \\
\text { Let } g(t)=t^{3}+2 t . \text { Then the inequality is } \\
g\left(\frac{1}{x^{2}}\right)>g\left(x^{2}+1\right) .
\end{array}
$$

Clearly, $g(t)=t^{3}+2 t$ is an increasing function on $\mathbf{R}$. Therefore, the above inequality is equivalent to $\frac{1}{x^{2}}>x^{2}+1$, which is
$$
\left(x^{2}\right)^{2}+x^{2}-1<0 \text {. }
$$

Solving this, we get $x^{2}<\frac{\sqrt{5}-1}{2}$.
Thus, the solution set of the original inequality is $\left(-\sqrt{\frac{\sqrt{5}-1}{2}}, \sqrt{\frac{\sqrt{5}-1}{2}}\right)$.