Answer
$\cos \left( {\frac{1}{x}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin \left( {1/x} \right)}}{{{x^2}}}} dx \cr
& {\text{Let }}u = \frac{1}{x},{\text{ then }}du = - \frac{1}{{{x^2}}}dx,{\text{ }}\frac{1}{{{x^2}}}dx = - du \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int {\frac{{\sin \left( {1/x} \right)}}{{{x^2}}}} dx = \int {\sin u} \left( { - du} \right) \cr
& = - \int {\sin u} du \cr
& {\text{Integrate}} \cr
& = - \left( { - \cos u} \right) + C \cr
& = \cos u + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = \frac{1}{x} \cr
& = \cos \left( {\frac{1}{x}} \right) + C \cr} $$
