Integral of $\int{\frac{2x-3}{(3x^2-2x+4)\sqrt{x^2-3x+1}}}dx$

Question

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.

While researching I found a substitution like this: $x = \frac{az+b}{cz+d}$ but I was not useful. — Rolando González, Feb 11 '20 at 21:02
I asked Maple for an answer and it gave a rather horrific answer. — MasB, Feb 11 '20 at 21:14
Possibly what one would first do is to use an appropriate substitution to simultaneously remove the linear terms in the quadratic expressions, as I do in Sections 4, 5, 6 of my 2-part answer to How to integrate $ \int \frac{x}{\sqrt{x^4+10x^2-96x-71}}dx$? See Articles 291-312 (pp. 287-308) of A Treatise on the Integral Calculus by Joseph Edwards (1921). If you decide to fully carry out the integration, I suspect you'll be in for a pretty rough algebra ride. — Dave L. Renfro, Feb 11 '20 at 21:38

an4s · Answer 1 · 2020-02-11T22:48:58.223

$$\int\dfrac{2x - 3}{\left(3x^2 - 2x + 4\right)\sqrt{x^2 - 3x + 1}}\,\mathrm dx\equiv2\int\dfrac{2x - 3}{\left(3x^2 - 2x + 4\right)\sqrt{\left(2x - 3\right)^2 - 5}}\,\mathrm dx$$

Let $u = 2x - 3\implies\mathrm du = 2\,\mathrm dx$. Therefore,

$$\int\dfrac{2x - 3}{\left(3x^2 - 2x + 4\right)\sqrt{\left(2x - 3\right)^2 - 5}}\,\mathrm dx\equiv2\int\dfrac u{\left(3u^2 + 14u + 31\right)\sqrt{u^2 - 5}}\,\mathrm du$$

Let $u = \sqrt 5\sec(v)\implies \mathrm du = \sqrt5\sec(v)\tan(v)\,\mathrm dv$. Therefore,

$$\begin{align}\int\dfrac u{\left(3u^2 + 14u + 31\right)\sqrt{u^2 - 5}}\,\mathrm du&\equiv\int\dfrac{5\sec^2(v)\tan(v)}{\left(15\sec^2(v)+ 14\sqrt5\sec(v) + 31\right)\sqrt{5\sec^2(v) - 5}}\,\mathrm dv \\ &\stackrel{\sec^2(v) = 1 + \tan^2(v)}=\sqrt5\int\dfrac{\sec^2(v)}{15\sec^2(v) + 14\sqrt5\sec(v) + 31}\,\mathrm dv\end{align}$$

Perform tangent half-angle substitution.

$$\int\dfrac{\sec^2(v)}{15\sec^2(v) + 14\sqrt5\sec(v) + 31}\,\mathrm dv\equiv\int\dfrac{\left(1 + \tan^2\left(\frac v2\right)\right)^2}{\left(1 - \tan^2\left(\frac v2\right)\right)^2\left(\frac{15\left(1 + \tan^2\left(\frac v2\right)\right)^2}{\left(1 - \tan^2\left(\frac v2\right)\right)^2} + \frac{14\sqrt5\left(1 + \tan^2\left(\frac v2\right)\right)}{1 - \tan^2\left(\frac v2\right)} + 31\right)}\,\mathrm dv$$

Let $t = \tan\left(\dfrac v2\right)\implies\mathrm dv = \dfrac2{1 + t^2}\,\mathrm dt$. Therefore,

$$\begin{align}&\int\dfrac{\left(1 + \tan^2\left(\frac v2\right)\right)^2}{\left(1 - \tan^2\left(\frac v2\right)\right)^2\left(\frac{15\left(1 + \tan^2\left(\frac v2\right)\right)^2}{\left(1 - \tan^2\left(\frac v2\right)\right)^2} + \frac{14\sqrt5\left(1 + \tan^2\left(\frac v2\right)\right)}{1 - \tan^2\left(\frac v2\right)} + 31\right)}\,\mathrm dv\\&\equiv-\int\dfrac{1 + t^2}{\left(7\sqrt5 - 23\right)t^4 + 16t^2 - 7\sqrt5 - 23}\,\mathrm dt\end{align}$$

Now, factor the denominator and perform partial fraction decomposition.

score 0 · Answer 2 · answered Feb 11 '20 at 21:59

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.

Integral of $\int{\frac{2x-3}{(3x^2-2x+4)\sqrt{x^2-3x+1}}}dx$

2 Answers2