# Integrate $\int{\frac{x^2-1}{x^4+3x^3+5x^2+3x+1}}dx$

$$\int{\frac{x^2-1}{x^4+3x^3+5x^2+3x+1}}dx$$

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.

• Try $u=x+1/x$ to start with. Commented Sep 19, 2019 at 0:04
• @DonaldSplutterwit But I wonder how would you know you have to substitute that if you don't know the answer beforehand. Is that a kind of special integration technique? Commented Sep 19, 2019 at 0:29
• Observe that the polynomial in the denominator has coefficients that are "symmetrical" about the central term & then note the numerators is just right to make the substitution work. Commented Sep 19, 2019 at 0:32
• We can divide the denominator by $x^{4/2}$ as it's a reciprocal polynomial. See math.stackexchange.com/questions/480102/… Commented Sep 19, 2019 at 1:27
• @DonaldSplutterwit So, If the denominator's coefficients are not "symmetrical", this method wouldn't apply? Commented Sep 19, 2019 at 8:33

\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}
\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*}
• $$(1)$$ : substitute $$x=\dfrac{1-y}{1+y}$$
• $$(2\rm a,b)$$ : simplify and complete the square
• $$(3)$$ : substitute $$z=y^2+1$$