$\int\frac{x+\sqrt[3]{x^2}+\sqrt[6]{x}}{x(1+\sqrt[3]{x})}dx$ is equal to :

$\frac{3}{2}x^{2/3}+6\tan^{-1}x^{1/6}+C$ $\frac{3}{2}x^{2/3}-6\tan^{-1}x^{1/6}+C$ $\frac{-3}{2}x^{2/3}+6\tan^{-1}x^{1/6}+C$ None of these

$\frac{3}{2}x^{2/3}+6\tan^{-1}x^{1/6}+C$

Put $x=z^6⇒dx=6z^5dz$ $∴\int\frac{x+\sqrt[3]{x^2}+\sqrt[6]{x}}{x(1+\sqrt[3]{x})}dx=\int\frac{(z^6+z^4+z)6z^5\,dz}{z^6(1+z^2)}$ $=6\int\frac{z^5+z^3+z}{z^2+1}dz=6\int(z^3+\frac{1}{z^2+1})dz=\frac{3}{2}z^4+6\tan^{-1}z+C=\frac{3}{2}x^{2/3}+6\tan^{-1}x^{1/6}+C$