Answer
$$ - 2\cosh \left( {{x^{ - 1/2}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sinh \left( {{x^{ - 1/2}}} \right)}}{{{x^{3/2}}}}} dx \cr
& {\text{substitute }}u = {x^{ - 1/2}}{\text{ }} \cr
& du = - \frac{1}{2}{x^{ - 3/2}}dx \cr
& - 2du = \frac{1}{{{x^{3/2}}}}dx \cr
& = \int {\sinh \left( {{x^{ - 1/2}}} \right)\frac{1}{{{x^{3/2}}}}} dx \cr
& = \int {\sinh u\left( { - 2du} \right)} \cr
& = - 2\int {\sinh u} du \cr
& {\text{find the antiderivative }}\left( {{\text{formula 15 page 489}}} \right) \cr
& = - 2\cosh u + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {x^{ - 1/2}} \cr
& = - 2\cosh \left( {{x^{ - 1/2}}} \right) + C \cr} $$
