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One of my friends sent me a list of integrals (all without solutions ) one of those problems is: $$\int \frac{2020x^{2019}+2019x^{2018}+2018x^{2017}}{x^{4044}+2x^{4043}+3x^{4042}+2x^{4041}+x^{4040}+1}dx$$

the numerator is $\frac{d}{dx}x^{2018}(1+x+x^2)$ and the denominator is $\left(x^2\left(x^{2018}(1+x+x^2)\right)\right)^2 +1$ so unless I am mistaken this integral is in the form of $\int\frac{f'(x)}{1+(x^2f(x))^2}dx$ which I don't know how to solve, maybe (If this problem is unsolvable ) there is a typo, but I couldn't verify that whether this problem has a typo or not since wolfram alpha for some reason don't understand my input.

Since this question seems to be incorrect I wounder what is the result of $\int_{- \infty}^{\infty} \frac{2020x^{2019}+2019x^{2018}+2018x^{2017}}{x^{4044}+2x^{4043}+3x^{4042}+2x^{4041}+x^{4040}+1}dx$ Does this have a nice closed form ?

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  • $\begingroup$ only take $f(x)=x$,the answer is very complicated. $\endgroup$
    – Bowei Tang
    Commented May 24 at 7:20
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    $\begingroup$ Maybe the problem was about $\int_{-\infty}^{\infty}$ and maybe it was in a problem list on complex analyses? $\endgroup$
    – Sergei Nikolaev
    Commented May 24 at 7:27
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    $\begingroup$ @pie Were the previous problems about definite or about indefinite integrals? $\endgroup$
    – Sergei Nikolaev
    Commented May 24 at 7:32
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    $\begingroup$ This antiderivative is both elementary and non elementary at the same time in that it is the sum of elementary functions, but the sum is indexed over the roots of a transcendental equation. It is a living monstrosity of a counterexample that closed form is not beauty. $\endgroup$
    – Ninad Munshi
    Commented May 24 at 8:23
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    $\begingroup$ I suspect the problem's author wanted a solution of the form $\arctan[x^n(x^2+x+1)]+C$ but accidentally used one value of $n$ in the integrand's numerator with another in its denominator. $\endgroup$
    – J.G.
    Commented May 24 at 9:15

1 Answer 1

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As suggested in the comments by @Ninad Munshi, if we let $f(x)=x^{n}(1+x+x^2)$, putting $n=0$ (the trivial case) allows partial fraction decomposition of the integrand into: $$ \frac{1}{13}\frac{4x+7}{(x^2+1)} - \frac{1}{13} \frac{4x^5+15x^4+18x^3-x^2-22x-6}{x^6+2x^5+2x^4-x^2+1} = \frac{1}{13}\frac{4x+7}{(x^2+1)} - \frac{1}{13} \frac{P(x)}{Q(x)} $$ WolframAlpha solves it to: $$\frac{1}{13}\left(7\tan^{-1} (x) +2\ln(1+x^2) -\sum_{\omega_j: Q(\omega_j)=0 } \frac{P(\omega_j)}{Q'(\omega_j)}\ln (x-\omega_j) \right)$$ The interesting thing is that similar-looking solutions are there for the general case of $n≠0$. Only the coefficient of the $\arctan x$ term increases, and the polynomials $P$ and $Q$ grow in order.

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