$\cfrac{1}{{{x^{1/2}} + {x^{1/3}}}}$
[Hint : $\cfrac{1}{{{x^{1/2}} + {x^{1/3}}}} = \cfrac{1}{{{x^{1/3}}\left( {1 + {x^{1/6}}} \right)}},{\rm{Put}}\,x = {t^6}$ ]

: Let $I = \int {\cfrac{{dx}}{{{x^{1/2}} + {x^{1/3}}}}}$

L.C.M of 2 and 3 is 6
So Put $x = {t^6}$ $\Rightarrow$ $dx = 6{t^5}dt$

$\therefore$ $I = \int {\cfrac{{6{t^5}dt}}{{{t^3} + {t^2}}}} = 6\int {\cfrac{{{t^3}}}{{t + 1}}} dt$

Since the degree of numerator is greater than the denominator,

$\therefore$ the fraction is improper. First, we mark it proper by dividing ${t^3}$ by $t + 1$ .

$= 6\int {\left[ {{t^2} - t + 1 - \cfrac{1}{{t + 1}}} \right]dt = 6\left[ {\cfrac{{{t^3}}}{3} - \cfrac{{{t^2}}}{2} + t - \log \left| {t + 1} \right|} \right] + C}$

$= 2\sqrt x - 3{x^{1/3}} + 6{x^{1/6}} - 6\log \left| {{x^{1/6}} + 1} \right| + C$