Let $\frac{d}{dx}F\left(x\right)=\frac{e^{\sin x}}{x},$$\mathrm{x} > 0 . \mathrm{If}$ ${\int }_1^4\frac{2e^{\sin x^2}}{x}dx=F\left(k\right)-f\left(1\right)$ then one of the possible values of $k$ is ______.

$\frac{d}{dx}F\left(x\right)=\frac{e^{\sin x}}{x},$
and, ${\int }_1^4\frac{2e^{\sin x^2}}{x}dx=F\left(k\right)-f\left(1\right)$
L.H.S $={\int }_1^4\frac{2·e^{\sin x^2}}{x}dx$
Let$x^2=t\Rightarrow 2xdx=dt$
$I={\int }_1^{16}\frac{2e^{\sin t}}{\sqrt{t}}·\frac{dt}{2·\sqrt{t}}={\int }_1^{16}\frac{e^{\sin t}}{t}dt$
${\left.I={\int }_1^{16}\frac{df\left(t\right)dt}{dt}=F\left(t\right)\right]}_1^{16}$
$I=F\left(16\right)-F\left(1\right)$
And, R.H.S $I=F\left(k\right)-F\left(1\right)$

 Hence $K=1$