Answer
$$\frac{{13}}{{162}}$$
Work Step by Step
$$\eqalign{
& \int_4^9 {\frac{{x - \sqrt x }}{{{x^3}}}} dt \cr
& {\text{Split the numerator}} \cr
& = \int_4^9 {\left( {\frac{x}{{{x^3}}} - \frac{{\sqrt x }}{{{x^3}}}} \right)} dt \cr
& {\text{Writting }}\sqrt x {\text{ as }}{x^{1/2}} \cr
& = \int_4^9 {\left( {\frac{x}{{{x^3}}} - \frac{{{x^{1/2}}}}{{{x^3}}}} \right)} dx \cr
& use\,\,the\,\,rule\,\,\,\frac{{{x^u}}}{{{x^v}}} = {x^{u - v}} \cr
& = \int_4^9 {\left( {{x^{ - 2}} - {x^{ - 5/2}}} \right)} dx \cr
& {\text{using the power rule for integration}} \cr
& = \left. {\left( {\frac{{{x^{ - 1}}}}{{ - 1}} - \frac{{{x^{ - 3/2}}}}{{ - 3/2}}} \right)} \right|_4^9 \cr
& = \left. {\left( { - \frac{1}{x} + \frac{2}{{3{x^{3/2}}}}} \right)} \right|_4^9 \cr
& {\text{Using The Fundamental Theorem}} \cr
& = \left( { - \frac{1}{9} + \frac{2}{{3{{\left( 9 \right)}^{3/2}}}}} \right) - \left( { - \frac{1}{4} + \frac{2}{{3{{\left( 4 \right)}^{3/2}}}}} \right) \cr
& {\text{simplify}} \cr
& = \left( { - \frac{7}{{81}}} \right) - \left( { - \frac{1}{6}} \right) \cr
& = - \frac{7}{{81}} + \frac{1}{6} \cr
& = \frac{{13}}{{162}} \cr} $$
