[tex]\displaystyle\int\frac{\mathrm dx}{x^{1/2}+x^{1/3}}[/tex]
Let [tex]u=x^{1/6}[/tex], so that [tex]u^6=x[/tex] and [tex]6u^5\,\mathrm du=\mathrm dx[/tex]. Then [tex]x^{1/2}=x^{3/6}=u^3[/tex] and [tex]x^{1/3}=x^{2/6}=u^2[/tex].
[tex]\displaystyle\int\frac{6u^5}{u^3+u^2}\,\mathrm du=6\int\frac{u^3}{u+1}\,\mathrm du[/tex]
Then with [tex]t=u+1[/tex], so that [tex]u=t-1[/tex] and [tex]\mathrm du=\mathrm dt[/tex], we have
[tex]\displaystyle6\int\frac{(t-1)^3}t\,\mathrm dt=6\int\left(t^2-3t+3-\frac1t\right)\,\mathrm dt[/tex]
which should be easy to finish.
