<s>[INST] Evaluate $\int \frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1}dx$ using elementary methods I found this problem while doing some integration from my problem practice book (unkown name). It said to evaluate it using elementary methods.
Please help me evaluating the following integral using elementary methods
$$\int \frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1} \, dx.$$
I tried to factorize the denominator but it failed. I can't think of any substitution too. I tried to input this at wolfram alpha, and it showed the answer as a summation of a complex function formed using complex cube roots of unity.
Substitution of $x=\sec\theta$ also failed.
I am continuously thinking of this but I'm not getting how to start it. If I succeed I will post it. It will be great if someone could give a beautiful solution to this problem preferably using elementary methods.
I am a high school student in India. I know how to evaluate elementary integrals .
 [/INST] Presented below is a full solution via elementary integration.
The polynomial $x^4-2x^3-6x-1$ admits two real roots, $a=$ -0.1650 and $b=$ 2.8068, computed with the standard quartic-root formulae below
$$a,b=\frac12\left(1+\sqrt r \pm \sqrt{3-r+{14}{r^{-\frac12}}}\right)\tag1
$$
with $ r= 1+ \left(\frac43\right)^{\frac23}\left( \sqrt[3]{\sqrt{129}+9} - \sqrt[3]{\sqrt{129}-9} \right) $.
Thus, the denominator of the integrand factorizes as
$$x^4-2x^3-6x-1=(x-a)(x-b)[x^2+(a+b-2)x-(ab)^{-1}]
$$
leading to following partial fractionalization
\begin{align}
\frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1}
=1+ \frac {p}{x-a} + \frac {q}{x-b}+ \frac{(2-p-q )x-\frac1{ab}(2+\frac {p}a+ \frac {q}b)}{x^2+(a+b-2)x-\frac1{ab}}
\end{align}
with $p=\frac{a^3+a^2+3a-1}{2a^3-3a^2-3}$ and $q=\frac{b^3+b^2+3b-1}{2b^3-3b^2-3}$.
Then, integrate to obtain
\begin{align}
& \int \frac{(x^2-1)(x^2+3)}{x^4-2x^3-6x-1}dx\\
=& \>x +2p \ln|x-a| +2q \ln|x-b|+(1-p-q)\ln\left(x^2+(a+b-2)x-\frac1{ab}\right)\\
& \>-\frac{(1-p-q)(a+b-2)+\frac2{ab}(1+\frac pa + \frac qb)}{\sqrt{ -\frac{(a+b-2)^2}4-\frac1{ab}}}
\tan^{-1}\frac{x+\frac{a+b-2}2}{\sqrt{ -\frac{(a+b-2)^2}4-\frac1{ab}}}+C
\end{align}
The result, though rather involved in appearance, is expressed in terms of $a$ and $b$, the two real roots given in (1).
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