Integrate the function: $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$

Let $x = t^{6}$,then $dx = 6t^{5} dt$.
Substituting these into the integral:
$\int \frac{1}{x^{1/2} + x^{1/3}} dx = \int \frac{6t^{5}}{t^{3} + t^{2}} dt$
$= \int \frac{6t^{5}}{t^{2}(t + 1)} dt = 6 \int \frac{t^{3}}{t + 1} dt$
By performing polynomial division,$\frac{t^{3}}{t + 1} = t^{2} - t + 1 - \frac{1}{t + 1}$.
So,the integral becomes:
$6 \int (t^{2} - t + 1 - \frac{1}{t + 1}) dt = 6 [\frac{t^{3}}{3} - \frac{t^{2}}{2} + t - \log|t + 1|] + C$
$= 2t^{3} - 3t^{2} + 6t - 6 \log|t + 1| + C$
Since $t = x^{1/6}$,substituting back gives:
$= 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6 \log|x^{1/6} + 1| + C$