Integrating $ \int\frac{1}{\sqrt{x}+\sqrt[3]{x}}\, dx$ I'm stuck on this integration with substitution problem:
$$
\int\frac{1}{\sqrt{x}+\sqrt[3]{x}}\,dx
$$
It tells me that I have to substitute with $\displaystyle u={x}^{\frac{1}{6}}$ but no matter what I try, I seem to either be getting further away from the answer of it is too long to compute analytically. I've tried multiplying by the conjugate, or factorising with the square root of $x$, but its not turning out how the answer is supposed to be.
Any help is much appreciated!!

Let 
\begin{eqnarray}
u &=& x^\frac{1}{6}\\
\Rightarrow x&=&u^6\\
\Rightarrow dx&=&6u^5 \cdot du 
\end{eqnarray}
Hence 
\begin{eqnarray}
\int\frac{1}{\sqrt{x}+\sqrt[3]{x}}dx&=& 6 \cdot \int\frac{u^5}{u^3+u^2}du\\
&=& 6 \cdot \int\frac{u^3}{u+1}du\\
&=& 6 \cdot \int\left(\frac{(u^3 + 1)}{u+1} - \frac{1}{u+1}\right)du\\
&=&6 \cdot \int(u^2 - u + 1) \cdot du - 6 \cdot \int\frac{1}{u+1} \cdot du\\
&=& 6 \cdot \left(\frac{u^3}{3} - \frac{u^2}{2} + u\right) - 6 \cdot \ln({u+1}) + C\\
&=& 6 \cdot \left(\frac{\sqrt{x}}{3} - \frac{\sqrt[3]{x}}{2} + \sqrt[6]{x} - \ln({\sqrt[6]{x}+1})\right) + C
\end{eqnarray}
$$\int\left(\frac{1}{\sqrt{x}+\sqrt[3]{x}}\right) \cdot dx = \boxed{\left(2 \sqrt{x} - 3 \sqrt[3]{x} + 6 \sqrt[6]{x} - 6 \ln({\sqrt[6]{x}+1})\right) + C}$$
