Answer
$$\frac{9}{{14}}$$
Work Step by Step
$$\eqalign{
& \int_1^e {\frac{1}{x}{{\left( {1 + 7\ln x} \right)}^{ - 1/3}}} dx \cr
& {\text{use substitution}}{\text{: }} \cr
& {\text{ }}u = 1 + 7\ln x,{\text{ so that }}\frac{{du}}{{dx}} = \frac{7}{x},\,\,\,dx = \frac{{xdu}}{7} \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = e,{\text{ }}u = 1 + 7\ln e = 8 \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ }}u = 1 + 7\ln 1 = 1 \cr
& {\text{then}} \cr
& \int_1^e {\frac{1}{x}{{\left( {1 + 7\ln x} \right)}^{ - 1/3}}} dx = \int_1^8 {\frac{1}{x}{u^{ - 1/3}}} \left( {\frac{{xdu}}{7}} \right) \cr
& = \frac{1}{7}\int_1^8 {{u^{ - 1/3}}} du \cr
& {\text{integrating, we get:}} \cr
& = \frac{1}{7}\left( {\frac{{{u^{2/3}}}}{{2/3}}} \right)_1^8 \cr
& = \frac{3}{{14}}\left( {{u^{2/3}}} \right)_1^8 \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{3}{{14}}\left( {{8^{2/3}} - {1^{2/3}}} \right) \cr
& {\text{simplifying, we get:}} \cr
& = \frac{3}{{14}}\left( {4 - 1} \right) \cr
& = \frac{9}{{14}} \cr} $$
