<s>[INST] Solve $\frac{1}{x-1}-\frac{4}{x-2}+\frac{4}{x-3}-\frac{1}{x-4}<\frac{1}{30} \forall x\in R$ I plugged this into WolframAlpha. It gave the factorized form as:
$$\frac{(x-7)(x-6)(x+1)(x+2)}{(x-4)(x-3)(x-2)(x-1)}>0$$
Once we've reached this point, I know how to find the solution for $x$. But, currently, I am having trouble converting the given question expression into the factorized form above. I started off strong with:
$$(\frac{1}{x-1}-\frac{1}{x-4})+(\frac{4}{x-3}-\frac{4}{x-2})<\frac{1}{30}$$
which simplifies to:
$$\frac{-3}{(x-1)(x-4)}+\frac{12}{(x-3)(x-2)}<\frac{1}{30}$$
which simplifies to:
$$\frac{3x^2-15x+10}{(x-1)(x-2)(x-3)(x-4)}<\frac{1}{90}$$
and now this is a dead end! As you can see, I got to the denomiator, but the numerator is not what I expected it to be. I could transfer $1/30$ to the LHS but that would bring the deadly fourth power into the numerator. If this was a competitive exam question, I would not be sitting in the hall factorizing fourth powers...

Is there a simpler method to factorize the LHS? Or is there an altogether different approach to solve this question?

 [/INST] Looking at those denominators, there seems to be some symmetry around $x=5/2$. Replacing $u=x-5/2$ we get
$$\frac{1}{u+\frac{3}{2}}-\frac{4}{u+\frac{1}{2}}+\frac{4}{u-\frac{1}{2}}-\frac{1}{u-\frac{3}{2}}$$
Now, $\frac{1}{u+B/2}-\frac{1}{u-B/2}=-\frac{4B}{4u^2-B^2}$ so the above reduces to
$$ -\frac{12}{4u^2-9}+\frac{16}{4u^2-1}$$
Letting know $s=4u^2 $ we have 
$$-\frac{12}{s-9}+\frac{16}{s-1} < \frac{1}{30}$$ which produces a quadratic inequation, which you should be able to solve - with the additional restriction $s\ge 0$. From that, you recover $x =  5/2 + u = (5\pm  \sqrt{s})/2 $ 
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