Answer
$$4\ln \left| {\frac{{{x^{1/4}}}}{{1 - {x^{1/4}}}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x\left( {1 - {x^{1/4}}} \right)}}} \cr
& {\text{Taking the hint on the book making a substitution of the form }}u = {x^{1/n}} \cr
& {\text{Take }}\,\,\,u = {x^{1/4}} \to \,\,\,\,\,\,\,\,{u^4} = x,\,\,\,\,\,\,\,4{u^3}du = dx\,\,\,\, \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{dx}}{{x\left( {1 - {x^{1/4}}} \right)}}} = \int {\frac{{4{u^3}du}}{{{u^4}\left( {1 - u} \right)}}} \cr
& = \int {\frac{{4du}}{{u\left( {1 - u} \right)}}} \cr
& \cr
& {\text{Integrate by tables using the formula }}\int {\frac{{du}}{{u\left( {a + bu} \right)}}} = \frac{1}{a}\ln \left| {\frac{u}{{a + bu}}} \right| + C \cr
& \int {\frac{{4du}}{{u\left( {1 - u} \right)}}} = 4\left( {\frac{1}{1}\ln \left| {\frac{u}{{1 - u}}} \right|} \right) + C \cr
& = 4\ln \left| {\frac{u}{{1 - u}}} \right| + C \cr
& \cr
& {\text{write the integrand in terms of }}x;{\text{ replace }}u = {x^{1/4}} \cr
& = 4\ln \left| {\frac{{{x^{1/4}}}}{{1 - {x^{1/4}}}}} \right| + C \cr} $$
