Answer
$$3\ln 5 + \frac{{12}}{{25}}$$
Work Step by Step
$$\eqalign{
& \int_1^5 {\left( {3{x^{ - 1}} + {x^{ - 3}}} \right)} dx \cr
& {\text{ integrate by using }}\int_a^b {{x^n}dx} = \left( {\frac{{{x^{n + 1}}}}{{n + 1}}} \right)_a^b{\text{ and }}\int_a^b {\frac{1}{x}} dx = \left. {\ln \left| x \right|} \right|_a^b + C \cr
& = \left( {3\ln \left| x \right| + \frac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right)_1^5 \cr
& = \left( {3\ln \left| x \right| + \frac{{{x^{ - 2}}}}{{ - 2}}} \right)_1^5 \cr
& = \left( {3\ln \left| x \right| - \frac{1}{{2{x^2}}}} \right)_1^5 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( {3\ln \left| 5 \right| - \frac{1}{{2{{\left( 5 \right)}^2}}}} \right) - \left( {3\ln \left| 1 \right| - \frac{1}{{2{{\left( 1 \right)}^2}}}} \right) \cr
& {\text{simplifying}} \cr
& = 3\ln 5 - \frac{1}{{50}} + \frac{1}{2} \cr
& = 3\ln 5 + \frac{{12}}{{25}} \cr} $$
