The answer of this integral
$$\int{\frac{x^2-1}{x^4+3x^3+5x^2+3x+1}}dx$$
is
$$\frac{2}{\sqrt{3}}\arctan\left[\frac{2}{\sqrt{3}}(x+\frac{1}{x})+\sqrt{3}\right]+C$$
But, I couldn’t figure out how I would solve it. I tried to use partial fraction, but the denominator
$$x^4+3x^3+5x^2+3x+1$$
can't be easily factored.
I also tried to use WolframAlpha to solve it, but it can't give a useful answer for this integral.
On
$$\begin{align*} I &= \int \frac{x^2-1}{x^4+3x^3+5x^2+3x+1} \, dx \\ &= -2 \int \frac{\frac{(1-y)^2}{(1+y)^2}-1}{\frac{(1-y)^4}{(1+y)^4} + 3\frac{(1-y)^3}{(1+y)^3} + 5\frac{(1-y)^2}{(1+y)^2} + 3\frac{1-y}{1+y} + 1} \, \frac{dy}{(1+y)^2} \tag1 \\ &= 8 \int \frac y{y^4+2y^2+13} \, dy \tag{2a} \\ &= 8 \int \frac y{\left(y^2 + 1\right)^2 + 12} \, dy \tag{2b} \\ &= 4 \int \frac{dz}{z^2+12} \tag3 \end{align*}$$
\begin{align} &\int{\frac{x^2-1}{x^4+3x^3+5x^2+3x+1}}dx\\ =&\int \frac{ 1 - \frac{1}{x^2}}{x^2+3x+5+\frac{3}{x}+\frac{1}{x^2}}dx =\int \frac{ d\left( x+\frac{1}{x}\right) }{\left( x+\frac{1}{x}+\frac{3}{2}\right)^2 + \frac 34 } \\ =&\ \frac{2}{\sqrt{3}}\arctan\left[\frac{2}{\sqrt{3}}\left(x+\frac{1}{x}\right)+\sqrt{3}\right]+C \end{align}