<s>[INST] 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? 
 [/INST] 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:
$$(2x-1)^\frac{1}{2}+2\ln (|(2x-1)^\frac{1}{4}-1|) +2(2x-1)^\frac{1}{4}+C$$
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