Let $$\frac{{df\left( x \right)}}{{dx}} = \frac{{{e^{\sin \,x}}}}{x},\,x > 0.$$     If $$\int_1^4 {\frac{{3{e^{\sin \,{x^3}}}}}{x}dx = f\left( k \right) - f\left( 1 \right)} $$       then one of the possible values of $$k$$ is :

Solution :
$$\eqalign{ & \frac{{df\left( x \right)}}{{dx}} = \frac{{{e^{\sin \,x}}}}{x} \cr & \Rightarrow \frac{{df\left( {{x^3}} \right)}}{{d\left( {{x^3}} \right)}} = \frac{{{e^{\sin \,{x^3}}}}}{{{x^3}}} \cr & \Rightarrow \frac{{df\left( {{x^3}} \right)}}{{d\left( {{x^3}} \right)}}.\frac{{d\left( {{x^3}} \right)}}{{dx}} = \frac{{{e^{\sin \,{x^3}}}}}{{{x^3}}}.3{x^2} \cr & \Rightarrow \frac{{df\left( {{x^3}} \right)}}{{dx}} = \frac{{3{e^{\sin \,{x^3}}}}}{x} \cr & \Rightarrow \int_1^4 {\frac{{3{e^{\sin \,{x^3}}}}}{x}dx} = \left[ {f\left( {{x^3}} \right)} \right]_{x = 1}^4 = f\left( {64} \right) - f\left( 1 \right) \cr} $$