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