$\begin{array}{l}
I = \int {\dfrac{{dx}}{{\sqrt x  + \sqrt[3]{x}}}} \\
\sqrt[6]{x} = t \Rightarrow \left\{ \begin{array}{l}
\sqrt x  = {t^3}\\
\sqrt[3]{x} = {t^2}\\
x = {t^6} \Rightarrow dx = 6{t^5}dt
\end{array} \right.\\
I = \int {\dfrac{{6{t^5}dt}}{{{t^3} + {t^2}}}}  = \\
\int {\dfrac{{6{t^3}dt}}{{t + 1}} = \int {\dfrac{{6\left( {{t^3} + 1} \right) - 6}}{{t + 1}}dt} } \\
 = \int {\dfrac{{6\left( {t + 1} \right)\left( {{t^2} - t + 1} \right) - 6}}{{t + 1}}dt} \\
 = \int {6\left( {{t^2} - t + 1} \right)dt}  - \int {\dfrac{6}{{t + 1}}dt} \\
 = 6\left( {\dfrac{{{t^3}}}{3} - \dfrac{{{t^2}}}{2} + t} \right) - 6\ln \left| {t + 1} \right| + C\\
 = 6\left( {\dfrac{{\sqrt x }}{3} - \dfrac{{\sqrt[3]{x}}}{2} + \sqrt[6]{x}} \right) - 6\ln \left| {\sqrt[6]{x} + 1} \right| + C\\
 = 2\sqrt x  - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\ln \left( {\sqrt[6]{x} + 1} \right) + C
\end{array}$

`int(dx)/(sqrtx+root[3]{x}`$\\$ `=\int(dx)/(x^(1/2).x^(1/3))=int(dx)/x^(5/6)=intx^(-5/6)dx`$\\$ `=6x^(1/6)+C`