Difficult Integration $\frac{1}{(2x-1)^{\frac {1}{2}}-(2x-1)^{\frac{1}{4}}}$ I am trying to solve this integral:$$\int{\frac{1}{(2x-1)^{\frac {1}{2}}-(2x-1)^{\frac{1}{4}}}}dx$$
My idea was to replace the second term in the denominator as $u$ and therefore have $\int{\frac{1}{u^2-u}}dx$ but it looks still complicated. On WolframAlpha I have  a long way to solve it. Do you have any advice of a shorter approach? 
 A: Hint
Considering the integral $$I=\int{\frac{dx}{(2x-1)^{\frac {1}{2}}-(2x-1)^{\frac{1}{4}}}}$$ make a change of variable $2x-1=u^4$ to get rid of the radicals; so $x=\frac{1}{2} \left(u^4+1\right)$, $dx=2 u^3\, du$ and so $$I=2\int \frac{u^2}{u-1}\,du$$ Now, change again variable $u=v+1$, $du=dv$ which makes $$I=2\int \frac{ (v+1)^2}{v}\,dv$$ and expand the numerator and divide each term by $v$. You will obtain very simple integrals.
I am sure that you can take from here.
A: Since setting $u=(2x-1)^{1/4}$ gives us $dx=2u^3du$, we have
$$\begin{align}\int\frac{1}{(2x-1)^{1/2}-(2x-1)^{1/4}}dx&=\int\frac{2u^3du}{u^2-u}\\&=\int\frac{2u^2}{u-1}du\\&=\int\left(2u+2+\frac{2}{u-1}\right)du\\&=u^2+2u+2\ln|u-1|+C\\&=\sqrt{2x-1}+2(2x-1)^{\frac 14}+2\ln|(2x-1)^{\frac 14}-1|+C\end{align}$$
A: Since I realized that I think I'm nearly done with a solution, here's the continuation:
For the record, with that $u$-substitution, the integral becomes:
$$u=(2x-1)^\frac{1}{4} \rightarrow x=\frac{u^4+1}{2}$$
$$du=\frac{1}{4}(2x-1)^\frac{-3}{4}2dx$$
$$\int \frac{1}{u^2-u}dx$$
$$\int \frac{1}{u^2-u}\left( 2(2x-1)^\frac{3}{4} \right)du$$
$$\int \frac{1}{u^2-u}\left( 2(u^4)^\frac{3}{4} \right)du$$
$$\int \frac{1}{u^2-u}\left( 2u^3 \right)du$$
$$\int \frac{2u^3}{u^2-u}du$$
$$\int \frac{2u^2}{u-1}du$$
Do polynomial division:
$$\int \left ( 2u+\frac{2}{u-1}+2 \right) du$$
This is trivial to integrate:
$$u^2+2\ln (|u-1|) +2u+C$$
Substitute back to the original $x$ variable:
$$(2x-1)^\frac{1}{2}+2\ln (|(2x-1)^\frac{1}{4}-1|) +2(2x-1)^\frac{1}{4}+C$$
