New Integral

Since the equation is even on both sides, we only need to consider for $x \ge 0$ to find the positive solution $x_2$ and then $x_1 = - x_2$ , since $x_1 < x_2$ . For $0 \le x \le 1$ ,

$\begin{aligned} 2\left(x^2 + \frac 1{x^2} \right) + 1 - x^2 & = 4 \left(\frac 32 - 2^{x^2-3} - \frac 1{2^{x^2+1}} \right) \\ x^2 + \frac 2{x^2} + 1 & = 6 - 2^{x^2-1} - \frac 1{2^{x^2-1}} \\ x^2 + \frac 2{x^2} + 2^{x^2-1} + \frac 1{2^{x^2-1}} & = 5 \end{aligned}$

By inspection, the solution is $x^2 = 1$ or $x_1 = -1$ and $x_2 = 1$ . Then

$\begin{aligned} I & = \int_0^4 \left \{\frac x4 \right \} \left(1 + \left \lfloor \tan \left(\frac {\{x\}}{1+\{x\}} \right) \right \rfloor \right) dx & \small \blue{\text{Note that }\left \lfloor \tan \left(\frac {\{x\}}{1+\{x\}} \right) \right \rfloor = 0} \\ & = \int_0^4 \left \{\frac x4 \right \} dx = \int_0^4 \frac x4\ dx = \frac {x^2}8 \ \bigg|_0^4 = \boxed 2 \end{aligned}$