<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] This is correct.  As a check: \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}u} & 6 \left( \frac{u^3}{3} + \frac{u^2}{2} + u + \ln |u-1| \right)  \\
  &= 6 \left( u^2 + u + 1 + \frac{1}{u-1} \right)  \\
  &= \frac{6u^3}{u-1}  \text{,}
\end{align*}
as expected.
It can be a little easier to start with $\mathrm{lcm}(2,3) = 6$, so 
$$  \frac{1}{x^{1/2} - x^{1/3}} = \frac{1}{x^{3/6} - x^{2/6}}  $$
and $u = x^{1/6}$, so both $x^{5/6} = u^5$ and $\mathrm{d}u = \frac{1}{6x^{5/6}} \,\mathrm{d}x$, and we conclude $\mathrm{d}x = 6u^5 \,\mathrm{d}u$.  Substituting into the integral leaves
$$  \int \frac{6 u^5 \,\mathrm{d}u}{u^3 - u^2} = \int \frac{6 u^3 \,\mathrm{d}u}{u - 1}  \text{,}  $$
and continue as you did.
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