Answer
$$6\root 6 \of x + 3\ln \left| {\frac{{\root 6 \of x - 1}}{{\root 6 \of x + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\left( {{x^{1/3}} - 1} \right)\sqrt x }}} dx \cr
& {\text{Take the hint }}x = {u^6},\,\,\,\,dx = 6{u^5}du \cr
& {\text{Write the integrand in terms of }}u \cr
& = \int {\frac{1}{{\left( {{u^2} - 1} \right){u^3}}}} \left( {6{u^5}} \right)du \cr
& = \int {\frac{{6{u^2}}}{{{u^2} - 1}}} du \cr
& {\text{Use long division }} \cr
& \frac{{6{u^2}}}{{{u^2} - 1}} = 6 + \frac{6}{{{u^2} - 1}} \cr
& \cr
& {\text{Decompose }}\frac{6}{{{u^2} - 1}}{\text{ into partial fractions}} \cr
& \frac{6}{{{u^2} - 1}} = \frac{6}{{\left( {u + 1} \right)\left( {u - 1} \right)}} \cr
& \frac{6}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = \frac{A}{{u + 1}} + \frac{B}{{u - 1}} \cr
& {\text{Multiply by }}\left( {u + 1} \right)\left( {u - 1} \right){\text{ and simplify}} \cr
& 6 = A\left( {u - 1} \right) + B\left( {u + 1} \right) \cr
& {\text{if }}u = - 1,\,\,\,A = - 3 \cr
& {\text{if }}u = 1,\,\,\,B = 3 \cr
& \cr
& {\text{Replace the coefficients}} \cr
& \frac{6}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = \frac{{ - 3}}{{u + 1}} + \frac{3}{{u - 1}} \cr
& \cr
& \int {\left( {6 + \frac{6}{{{u^2} - 1}}} \right)du} = \int 6 du + \int {\left( {\frac{{ - 3}}{{u + 1}} + \frac{3}{{u - 1}}} \right)du} \cr
& {\text{Integrate}} \cr
& = 6u - 3\ln \left| {u + 1} \right| + 3\ln \left| {u - 1} \right| + C \cr
& = 6u + 3\ln \left| {\frac{{u - 1}}{{u + 1}}} \right| + C \cr
& {\text{Replace }}u = \root 6 \of x \cr
& = 6\root 6 \of x + 3\ln \left| {\frac{{\root 6 \of x - 1}}{{\root 6 \of x + 1}}} \right| + C \cr} $$
