Answer
$$\ln 4 - 7$$
Work Step by Step
$$\eqalign{
& \int_1^8 {\left( {\frac{2}{{3x}} - \frac{8}{{{x^2}}}} \right)} dx \cr
& {\text{use the sum rule}} \cr
& = \int_1^8 {\frac{2}{{3x}}} dx - \int_1^8 {\frac{8}{{{x^2}}}} dx \cr
& = \frac{2}{3}\int_1^8 {\frac{1}{x}} dx - 8\int_1^8 {{x^{ - 2}}} dx \cr
& {\text{integrate using the power rule and }}\cr
&\int {\frac{1}{x}} dx = \ln \left| x \right| + C \cr
& = \frac{2}{3}\left( {\ln \left| x \right|} \right)_1^8 - 8\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right)_1^8 \cr
& = \left( {\frac{2}{3}\ln \left| x \right| + \frac{8}{x}} \right)_1^8 \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \left( {\frac{2}{3}\ln \left| 8 \right| + \frac{8}{8}} \right) - \left( {\frac{2}{3}\ln \left| 1 \right| + \frac{8}{1}} \right) \cr
& {\text{simplifying, we get:}} \cr
& = \ln 4 + 1 - 8 \cr
& = \ln 4 - 7 \cr} $$
