Let $\frac{d}{dx}F(x)=\frac{e^{\sin x}}{x}$, $x>0$. If ${\int }_1^4\frac{3e^{\sin x^3}}{x}dx=F(k)-F(1)$, then one of the possible values of $k$, is:

Let $I={\int }_1^4\frac{3e^{\sin x^3}}{x}dx$.Substitute $u=x^3⟹du=3x^{2}dx⟹\frac{du}{3x^2}=dx$.Also $\frac{du}{3u}=\frac{dx}{x}$.The integral becomes:
$$I={\int }_{1^3}^{4^3}3e^{\sin u}\frac{du}{3u}={\int }_1^{64}\frac{e^{\sin u}}{u}du$$
We are given $\frac{d}{dx}F(x)=\frac{e^{\sin x}}{x}$, so the antiderivative of $\frac{e^{\sin u}}{u}$ is $F(u)$.
$$I=[F(u){]}_1^{64}=F(64)-F(1)$$
Comparing with $F(k)-F(1)$, we get $k=64$.