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I need :

$$ \int \frac{\mathrm dx}{\sqrt{x}(1+\sqrt[3]{x})}$$

Can you give me a hint, which way would suit best? I tried substitution, but don't have an idea what to substitute.. :/

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HINT:

As the Integrand involves $\displaystyle \sqrt x=x^{\frac12},\sqrt[3]x=x^{\frac13}$

and $\displaystyle\mathrm{gcd}\left(\frac12,\frac13\right)=\frac1{\mathrm{lcm}(2,3)}=\frac16$

put $\displaystyle y=x^{\frac16}\implies \sqrt x=y^3,\sqrt[3]x=y^2$ and $\displaystyle x=y^6$

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$$ \int\frac{dx}{\sqrt x\left(1+\sqrt[3]{x}\right)}=\int\frac{dx}{\sqrt[6]x^{3}\left(1+\sqrt[6]{x^2}\right)}=|\sqrt[6]{x}=t\Rightarrow x=t^6\Rightarrow dx=6t^5dt| $$ $$ =\int\frac{6t^5dt}{t^3\left(1+t^2\right)}=6\int\frac{t^2dt}{1+t^2}=6\int\frac{t^2+1-1}{1+t^2}dt=6\int\left(1-\frac{1}{1+t^2}\right)dt $$ $$ =6\int dt-6\int \frac{1}{1+t^2}dt=6t-6\arctan t=6(\sqrt[6]{x}-\arctan \sqrt[6]{x})+C $$

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