Let $$\frac{d}{{dx}}F\left( x \right) = \frac{{{e^{\sin \,x}}}}{x},\,x > 0.$$      If $$\int_1^4 {\frac{{2{e^{\sin \,{x^2}}}}}{x}dx} = F\left( k \right) - F\left( 1 \right)$$       then one of the possible values of $$k$$ is :

Solution :
$$\eqalign{ & {\text{Put }}{x^2} = z \cr & {\text{Then }}\int_1^4 {\frac{{2.{e^{\sin \,{x^2}}}}}{x}dx} = \int_1^{16} {\frac{{{e^{\sin \,z}}}}{z}dz} \cr & = \int_1^{16} {\frac{d}{{dz}}\left\{ {F\left( z \right)} \right\}dz} \cr & = \left[ {F\left( z \right)} \right]_1^{16} \cr & = F\left( {16} \right) - F\left( 1 \right) \cr} $$
$$\therefore \,F\left( {16} \right) = F\left( k \right),$$     from the question.
Hence, one of the possible values of $$k=16.$$