Answer
$\frac{5}{4}\ln \left| {{x^{4/5}} - 1} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x - {x^{1/5}}}}} dx \cr
& {\text{Substitute }}x = {u^5},{\text{ then }}dx = 5{u^4}du \cr
& \int {\frac{1}{{x - {x^{1/5}}}}} dx = \int {\frac{1}{{{u^5} - {{\left( {{u^5}} \right)}^{1/5}}}}\left( {5{u^4}} \right)} du \cr
& = \int {\frac{{5{u^4}}}{{{u^5} - u}}} du \cr
& = \int {\frac{{5{u^3}}}{{{u^4} - 1}}} du \cr
& {\text{Rewrite the integrand}} \cr
& = \frac{5}{4}\int {\frac{{4{u^3}}}{{{u^4} - 1}}} du \cr
& {\text{Integrating}} \cr
& = \frac{5}{4}\ln \left| {{u^4} - 1} \right| + C \cr
& {\text{Write in terms of }}x,{\text{ }}x = {u^5} \to u = \root 5 \of x \cr
& = \frac{5}{4}\ln \left| {{{\left( {\root 5 \of x } \right)}^4} - 1} \right| + C \cr
& = \frac{5}{4}\ln \left| {{x^{4/5}} - 1} \right| + C \cr} $$
