Let $I = \int \frac{dx}{x^{1/2}+x^{1/3}}$ L.C.M. of 2 and 3 is 6So put $x = t^6 \Rightarrow dx = 6 t^5 dt$$\therefore I = \int \frac{6t^5 dt}{t^3+t^2}$ $= 6 \int \frac{t^3}{t+1} dt$Since the degree of numerator is greater than the denominator.∴ The fraction is improper. First we make it proper by dividing $t^3$ by $t + 1$$= 6 \int \left[ t^2 - t + 1 - \frac{1}{t+1} \right] dt$$= 6 \left[ \frac{t^3}{3} - \frac{t^2}{2} + t - \log|t+1| \right] + C$