Answer
$$\int \frac{1}{\sqrt{x}-\sqrt[3]{x}} d x =2 \sqrt{x}+3 \sqrt[3]{x}+6 \sqrt[6]{x}+6 \ln |\sqrt[6]{x}-1|+C $$
Work Step by Step
Given $$\int \frac{1}{\sqrt{x}-\sqrt[3]{x}} d x \ \ \ \ \ \ \ \ \ \ [\text { Hint: Substitute } u=\sqrt[6]{x} .]$$
\begin{array}{l}{\text { Let } u=\sqrt[6]{x} \text { . Then } x=u^{6}, \text { so } d x=6 u^{5} d u \\
\text { and } \sqrt{x}=u^{3}, \sqrt[3]{x}=u^{2} \text { . Thus, }} \\ {\qquad \begin{aligned} I=& \int \frac{d x}{\sqrt{x}-\sqrt[3]{x}} \\
&=\int \frac{6 u^{5} d u}{u^{3}-u^{2}}\\
&=6 \int \frac{u^{5}}{u^{2}(u-1)} d u\\
&=6 \int \frac{u^{3}}{u-1} d u \\
&=6 \int \frac{u^{3}-1+1}{u-1} d u \\
&=6 \int\left( \frac{u^{3}-1}{u-1}+\frac{ 1}{u-1}\right)d u \\
&=6 \int\left( \frac{(u -1)(u^2+u+1)}{u-1}+\frac{ 1}{u-1}\right)d u \\
&=6 \int\left(u^{2}+u+1+\frac{1}{u-1}\right) d u \\ &=6\left(\frac{1}{3} u^{3}+\frac{1}{2} u^{2}+u+\ln |u-1|\right)+C\\
&=2 \sqrt{x}+3 \sqrt[3]{x}+6 \sqrt[6]{x}+6 \ln |\sqrt[6]{x}-1|+C \end{aligned}}\end{array}