${\displaystyle \int \frac{\sqrt{x}}{\sqrt{x}+\sqrt[3]{x}}dx}$ is equal to

Put $x=z^6\Rightarrow dx=6z^{5}dz.$

$\therefore {\displaystyle \int \frac{\sqrt{x}}{\sqrt{x}+\sqrt[3]{x}}dx={\displaystyle \int \frac{z^{3}6z^{5}dz}{z^3+z^2}=6{\displaystyle \int \frac{z^{6}dz}{z+1}}}}$

$=6{\displaystyle \int \left(\frac{z^6-1}{z+1}+\frac{1}{z+1}\right)}dz$

$=6{\displaystyle \int \frac{\left(z^3-1\right)\left(z+1\right)\left(z^2-z+1\right)}{z+1}}dz+6{\displaystyle \int \frac{dz}{z+1}}$

$=6{\displaystyle \int \left(z^5-z^4+z^3-z^2+z-1\right)}dz+6\log \left(z+1\right)$

$=z^6-\frac{6}{z}{z}^5+\frac{3}{2}{z}^4-2z^3+3z^2-6z+6\log \left(z+1\right)+C$

$=x-\frac{6}{5}{x}^{5/6}+\frac{3}{2}{x}^{2/3}-2x^{1/2}+3x^{1/3}-6x^{1/6}+6\log \left(1+x^{1/6}\right)+C.$