Answer
\[ = \frac{4}{3}\ln \left| {{x^{\frac{3}{4}}} - 1} \right|+C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{x - \sqrt[4]{x}}}} \hfill \\
\hfill \\
set\,\,x = {u^4}\,\,then\,\,\,dx = 4{u^3}du \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{x - \sqrt[4]{x}}}} = \int_{}^{} {\frac{{4{u^3}}}{{{u^4} - u}}} \hfill \\
\hfill \\
\frac{{4{u^3}}}{{{u^4} - u}} = \frac{{4{u^3}}}{{u\left( {{u^3} - 1} \right)}} = \frac{{4{u^2}}}{{{u^3} - 1}} \hfill \\
\hfill \\
= 4\int_{}^{} {\frac{{{u^2}}}{{{u^3} - 1}}} du = \hfill \\
\hfill \\
factor\,\,the\,\,{\text{denominator}} \hfill \\
\hfill \\
4\int_{}^{} {\frac{{{u^2}}}{{\,\left( {u - 1} \right)\,\left( {{u^2} + u + 1} \right)}}du} \hfill \\
\hfill \\
Applying\,\,partial\,\,fractions\, \hfill \\
\hfill \\
= \frac{4}{3}\int_{}^{} {\,\,\left[ {\frac{1}{{u - 1}} + \frac{{2u + 1}}{{{u^2} + u + 1}}} \right]\,du} \hfill \\
\hfill \\
{\text{integrate}} \hfill \\
\hfill \\
= \frac{4}{3}\ln \left| {u - 1} \right| + \frac{4}{3}\left| {{u^2} + u + 1} \right|\, + C \hfill \\
\hfill \\
= \frac{4}{3}\ln \left| {{u^3} - 1} \right|+C \hfill \\
\hfill \\
substitute\,\,for\,\,\,u \hfill \\
\hfill \\
= \frac{4}{3}\ln \left| {{x^{\frac{3}{4}}} - 1} \right|+C \hfill \\
\hfill \\
\end{gathered} \]
