Question

If $x_{1}$ and $x_{2}\left(x_{1}< x_{2}\right)$ are two values of $x$ satisfying the equation $\left|2\left(x^{2}+\frac{1}{x^{2}}\right)+\right|\left|1-x^{2}\right|=4\left(\frac{3}{2}-2^{x^{2}-3}-\frac{1}{2^{x^{2}+1}}\right)$, then the value of $\int\limits_{x_{1}+x_{2}}^{3 x_{2}-x_{1}}\left\{\frac{x}{4}\right\}\left(1+\left[\tan \left(\frac{\{x\}}{1+\{x\}}\right)\right]\right) d x$, (where $[\cdot]$ and $[\cdot]$ denote greatest integer function and fractional part function respectively)

Answer 1

$\Rightarrow x=\pm 1 \Rightarrow x_{1}=-1$
$x_{2}=1$
$\int\limits_{0}^{4} \frac{x}{4} 1+\tan \frac{x}{1+x} d x$
$\int\limits_{0}^{4} \frac{x}{4} d x=\frac{1}{4} \frac{x^{2}}{2} 0$
$=\frac{1}{4}[8-0]=2$