## Calculus: Early Transcendentals (2nd Edition)

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