<s>[INST] Using partial fraction to power that less than 1 I need help to check this answer since, unfortunately, the worksheet doesn't attach the key.
Integrate:
$$\int \frac{1}{\sqrt{x} - \sqrt[3]{x}}dx$$
So, I tried use a substitution that allows the variables to become polynomials. I rewrote the expression to
$$\int \frac{1}{x^\frac{5}{6}\left(x^\frac{-2}{6}- x^\frac{-3}{6}\right)} \, dx$$
when I set $$u = x^\frac{1}{6}  \text{  and  } \frac{dx}{x^\frac{5}{6}}=6 \, du$$ then 
$$\int \frac{6 \, du}{\frac{1}{u^2}-\frac{1}{u^3}} = \int \frac{6 u^3 \, du}{u-1}.$$ 
This becomes
$$6 \int \left( u^2 + u + 1 + \frac{1}{u-1} \right) \, du = 6 \left(\frac{u^3}{3} + \frac{u^2}{2} + u + \ln\lvert u-1\rvert \right) + c.$$
and last part is to substitute back $u = x^\frac{1}{6}$. 
I don't know whether this right or not. This is first time I am dealing with partial factions that have less than one power. Please correct it if this work is wrong. Thanks in advance.
 [/INST] An alternate view it to do a power expansion and the cancel powers. In this example $u^3$ can be seen as:
$$u^3 = ( (u-1) + 1)^3 = (u-1)^3 + 3 \, (u-1)^2 + 3 \, (u-1) + 1$$
and 
$$\frac{u^3}{u-1} = (u-1)^2 + 3 \, (u-1) + 3 + \frac{1}{u-1}.$$
Integration yields:
\begin{align}
I &= \int \frac{u^3 \, du}{u-1} \\
&= \int (u-1)^2 \, du + 3 \, \int (u-1) \, du + 3 u + \int \frac{du}{u-1} \\
&= \frac{(u-1)^3}{3} + \frac{3 \, (u-1)^2}{2} + 3 \, u + \ln(u-1) + c_{0} \\
6 I &= 2 (u-1)^3 + 9 (u-1)^2 + 18(u-1) + 6 \, \ln(u-1) + c_{1}
\end{align}
This leads to the form
$$\int \frac{dx}{\sqrt{x} - \sqrt[3]{x}} = 2 (\sqrt[6]{x}-1)^3 + 9 (\sqrt[6]{x}-1)^2 + 18(\sqrt[6]{x}-1) + 6 \, \ln(\sqrt[6]{x}-1) + c_{1}. $$
</s>