Answer
$$\frac{2}{3}{\tan ^{ - 1}}{\left( {{x^3} - 1} \right)^{1/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\sqrt {{x^3} - 1} }}} dx \cr
& {\text{Taking the hint on the book making a substitution of the form }}u = {\left( {x + a} \right)^{1/n}} \cr
& {\text{Then}}{\text{,}} \cr
& {\text{Take }}\,\,\,u = {\left( {{x^3} - 1} \right)^{1/2}} \to \,\,\,\,\,\,\,\,{u^2} = {x^3} - 1,\,\,\,\,\,{x^3} = {u^2} + 1\,\,\,\, \cr
& 2udu = 3{x^2}dx,\,\,\,\,\,\,\,\,dx = \frac{{2udu}}{{3{x^2}}} \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{1}{{x\sqrt {{x^3} - 1} }}} dx = \int {\frac{1}{{x\left( u \right)}}} \left( {\frac{{2udu}}{{3{x^2}}}} \right) = \int {\frac{{2u}}{{3u{x^3}}}du} \cr
& = \int {\frac{{2u}}{{3u\left( {{u^2} + 1} \right)}}du} \cr
& = \frac{2}{3}\int {\frac{1}{{{u^2} + 1}}du} \cr
& \cr
& {\text{integrating}} \cr
& = \frac{2}{3}{\tan ^{ - 1}}u + C \cr
& \cr
& {\text{write the integrand in terms of }}x;{\text{ replace }}u = {\left( {{x^3} - 1} \right)^{1/2}} \cr
& = \frac{2}{3}{\tan ^{ - 1}}{\left( {{x^3} - 1} \right)^{1/2}} + C \cr} $$
