1

I am required to find the following (Source is unknown):$$\int_{-5}^{-2} (\frac{x^2-x}{x^3-3x+1})^2 dx+\int_{\frac{1}{6}}^{\frac{1}{3}} (\frac{x^2-x}{x^3-3x+1})^2 dx+\int_{\frac{6}{5}}^{\frac{3}{2}} (\frac{x^2-x}{x^3-3x+1})^2 dx$$

Initial thoughts:

The integrands are same but with different limits. The limits are related to each other in a very noticeable way, however I am not able to exploit his. Like: $-1* \frac{1}{-2} (\text{reciprocal of the limit of the first integral})* \frac{1}{\frac{1}{3}}(\text{reciprocal of the limit of the second integral}) = \frac{3}{2}(\text{ limit of the third integral})$,

Some tries:

Tried to solve: $\int (\frac{x^2-x}{x^3-3x+1})^2 dx$

$x^3-3x+1$ invites a substitution like $x=2\cos{t}$, since it could be expressed as $\cos{3t}$. However I felt it was of no use. Wolfram couldn't integrate it in terms of elementary functions which suggests, this probably has something to do with definite integral properties, and some clever trick which I am not able to find.

Maybe the integrand is of the form $\frac{d}{dx}(\frac{f(x)}{x^3-3x+1})$ for some $f(x)$ but I was unsuccessful in finding such a $f(x)$ either.

For $\int_{\frac{1}{6}}^{\frac{1}{3}} (\frac{x^2-x}{x^3-3x+1})^2 dx$ I tried things like $x \to \frac{1}{t}$ but it didnt really budge. I could change all the limits of the integrals to $0$ and $1$, however the difference of the upper and lower limits for each of the integrals dont really look nice either.

q123LsaB
  • 327
  • 1
  • 12
  • I'm quite sure that the integral of nonnegative function is not $-\frac{2}{27}$; in fact it seems You skipped quite a big part of the Wolfram's answer – Wham Bang Shang-a-Lang Apr 25 '23 at 19:26
  • @WhamBangShang-a-Lang Extremely sorry. Ill cut the part out. – q123LsaB Apr 25 '23 at 19:27
  • Nothing to be sorry about. Integral calculator here: https://mathdf.com/int/ provides a step-by-step solution for each of the integrals, but it looks terrible and clearly not a proper approach. I'll just throw in the observation that not only the limits multiply to $-1$ (as You've pointed out), but also the roots of the polynomial in the denominator – Wham Bang Shang-a-Lang Apr 25 '23 at 19:54
  • https://math.stackexchange.com/questions/3104918/evaluating-int-5-2-fx-dx-int-1-61-3fx-dx-int-6-53-2?noredirect=1 – Accelerator Apr 26 '23 at 00:12
  • @Accelerator Thank you. So the correct idea was to try to convert to a single limit. I'm new to stackexchange, could you tell me exactly how to search for questions? – q123LsaB Apr 26 '23 at 05:12
  • 1
    I use Approach Zero. It works most of the time. – Accelerator Apr 26 '23 at 05:45

1 Answers1

1

Just for the fun

The antiderivative is not so complicated and does not require any change of variable.

Write $$\frac{x^2-x}{x^3-3 x+1}=\frac{x(x-1)}{(x-a) (x-b) (x-c)}$$ where $$a=2 \cos \left(\frac{2 \pi }{9}\right) \qquad \qquad b=2 \sin \left(\frac{\pi }{18}\right)\qquad\qquad c=-2 \cos \left(\frac{\pi }{9}\right)$$ are the roots of the denominator.

Using partial fraction decomposition $$\frac{x^2-x}{x^3-3 x+1}=\frac A{x-a}+\frac B{x-b}+\frac C{x-c}$$ where $$A=\frac{(a-1) a}{(a-b) (a-c)}\qquad B=\frac{(b-1) b}{(b-a) (b-c)} \qquad C=\frac{(c-1) c}{(c-a) (c-b)}$$

Squaring, the integrand is $$\frac{\left(x^2-x\right)^2}{\left(x^3-3 x+1\right)^2}=\frac{A^2}{(x-a)^2}+\frac{B^2}{(x-b)^2}+\frac{C^2}{(x-c)^2}+$$ $$\frac{2 A B}{(x-a) (x-b)}+\frac{2 A C}{(x-a) (x-c)}+\frac{2 B C}{(x-b) (x-c)}$$ We use again partial fraction decomposition for the last three pieces. Integrating termwise $$\int\frac{\left(x^2-x\right)^2}{\left(x^3-3 x+1\right)^2}\,dx=\frac{A^2}{a-x}+\frac{B^2}{b-x}+\frac{C^2}{c-x}+$$ $$\frac{2 A (a (B+C)-b C-B c)}{(a-b) (a-c)}\,\log (a-x)+$$ $$\frac{2 B (C (a-b)+A (c-b))}{(a-b) (b-c)}\, \log (b-x)+$$ $$\frac{2 C (B (a-c)+A (b-c))}{(a-c) (c-b)}\,\log (c-x)$$

Now, tedious calculations and simplifications to arrive at a total of $\frac{624}{109}$ as given here.

Claude Leibovici
  • 260,315