Let  $\frac{d}{dx}\left(F\right(x\left)\right)=\frac{e^{\sin x}}{x},x>0$ 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, 

Solution:

We have,

$\frac{d}{dx}\left(F\right(x\left)\right)=\frac{e^{\sin x}}{x}\Rightarrow \int \frac{e^{\sin x}}{x}dx=F\left(x\right)$               …(i)

Now, 

${\int }_1^4\frac{e^{\sin x^2}}{x}dx={\int }_1^4 \frac{e^{\sin x^2}}{x^2}\cdot d\left(x^2\right)={\int }_1^{16} \frac{e^{\sin t}}{t}dt$, where $t=x^2$

${\int }_1^{4}2\frac{e^{\sin x^2}}{x}dx=\left[F\right(t){]}^{16}=F(16)-F(1)$      [using (i)]

hence , $k=16$