$\frac{d}{dx}F\left(x\right)=\frac{e^{\sin x}}{x},x>0.If{\displaystyle \underset{1}{\overset{4}{\int }}\frac{2e^{\sin x^2}}{x}dx=F\left(k\right)-F\left(1\right),}$ then one of the possible values of k is :

$F\left(x\right)={\displaystyle \int \frac{e^{Sinx}}{x}dx}$

            =${\displaystyle \underset{1}{\overset{4}{\int }}\frac{2.e^{Sin{x}^2}}{x}dx}$

           $={\displaystyle \underset{1}{\overset{4}{\int }}\frac{2x.e^{Sin{x}^2}}{x^2}dx}$        $\begin{array}{c}{x}^2=t,\\ 2xdx=dt\end{array}$

           $={{\displaystyle \underset{1}{\overset{16}{\int }}\frac{e^{Sint}}{t}dt=\left(F\left(t\right)\right)}}_1^{16}$

           $=f\left(16\right)-F\left(1\right)$

$\therefore k=16$