<s>[INST] Integral of $\int{\frac{2x-3}{(3x^2-2x+4)\sqrt{x^2-3x+1}}}dx$ I need to solve the following integral:
$$\int{\frac{2x-3}{(3x^2-2x+4)\sqrt{x^2-3x+1}}}dx$$
First I tried factorizing the denominators but it wasn't helpful. Then I thought about completing the square in $3x^2-2x+4$ but nothing came from there. I don't know if there might be any useful trigonometric substitution I could use or maybe some other substitution. I also tried with $t = \sqrt{x^2-3x+1}$  and $t = \frac{1}{x^2-2x+4}$ but also got me nowhere. Thanks for reading and your help. 
 [/INST] Using the substitution $u^2 = x^2-3x+1$ we get that
$$2du = \frac{2x-3}{\sqrt{x^2-3x+1}}dx$$
which is exactly $2$ out of the $3$ terms that we need to kill for our integral. Then using quadratic equation to solve for $x$ we get that
$$x = \frac{3}{2}\pm\sqrt{u^2+\frac{5}{4}}$$
$$x^2 = u^2 + \frac{7}{3} \pm 3\sqrt{u^2+\frac{5}{4}}$$
so our integral becomes the slightly nicer looking
$$\int \frac{2\:du}{3u^2+8\pm 7\sqrt{u^2+\frac{5}{4}}}$$
and at this point so far trig and hyperbolic substitutions haven't been working nicely. Wolfram is not giving a straight answer, either, so who knows if an elementary real answer exists.
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