Answer
$$\frac{{108}}{{25}}$$
Work Step by Step
$$\eqalign{
& \int_1^5 {\left( {6{n^{ - 2}} - {n^{ - 3}}} \right)} dn \cr
& {\text{integrate by using }}\int_a^b {{x^y}} dy = \left( {\frac{{{x^{y + 1}}}}{{y + 1}}} \right)_a^b \cr
& = \left( {6\left( {\frac{{{n^{ - 2 + 1}}}}{{ - 2 + 1}}} \right) - \left( {\frac{{{n^{ - 3 + 1}}}}{{ - 3 + 1}}} \right)} \right)_1^5 \cr
& = \left( {6\left( {\frac{{{n^{ - 1}}}}{{ - 1}}} \right) - \left( {\frac{{{n^{ - 2}}}}{{ - 2}}} \right)} \right)_1^5 \cr
& = \left( { - 6{n^{ - 1}} + \frac{{{n^{ - 2}}}}{2}} \right)_1^5 \cr
& {\text{use }}\frac{1}{{{x^a}}} = {x^{ - a}} \cr
& = \left( { - \frac{6}{n} + \frac{1}{{2{n^2}}}} \right)_1^5 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( { - \frac{6}{5} + \frac{1}{{2{{\left( 5 \right)}^2}}}} \right) - \left( { - \frac{6}{1} + \frac{1}{{2{{\left( 1 \right)}^2}}}} \right) \cr
& = - \frac{{59}}{{50}} - \left( { - \frac{{11}}{2}} \right) \cr
& = - \frac{{59}}{{50}} + \frac{{11}}{2} \cr
& = \frac{{108}}{{25}} \cr} $$
