Is there a quicker way to evaluate $\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$? The integral is: $$\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$$
My procedure:
$$4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3)^2-5}\ dx=4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3-\sqrt5)(2x^2+3+\sqrt5)}\ dx$$
$$$$Using partial fractions to separate the integral, we obtain:
$$\frac1{\sqrt5}\int_{0}^{\infty} \left(\frac{5-\sqrt5}{2x^2+3-\sqrt5}-\frac{5+\sqrt5}{2x^2+3+\sqrt5}\right)\ dx$$ 
$$=\frac{5-\sqrt5}{\sqrt5}\int_{0}^{\infty} \frac1{2x^2+3-\sqrt5}\ dx-\frac{5+\sqrt5}{\sqrt5}\int_{0}^{\infty}\frac1{2x^2+3+\sqrt5}\ dx$$
$$$$We need write the fractions in the form $\frac1{u^2+1}$:
$$=\frac{5-\sqrt5}{\sqrt5}\int_{0}^{\infty} \frac1{(3-\sqrt5)\left(\frac{2x^2}{3-\sqrt5}+1\right)}\ dx-\frac{5+\sqrt5}{\sqrt5}\int_{0}^{\infty}\frac1{(3+\sqrt5)\left(\frac{2x^2}{3+\sqrt5}+1\right)}\ dx$$
$$=\frac{5-\sqrt5}{\sqrt5(3-\sqrt5)}\int_{0}^{\infty} \frac1{\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)^2+1}\ dx-\frac{5+\sqrt5}{\sqrt5(3+\sqrt5)}\int_{0}^{\infty}\frac1{\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)^2+1}\ dx$$
$$\frac{5-\sqrt5}{\sqrt{10(3-\sqrt5)}}\int_{0}^{\infty} \frac{\sqrt2}{\sqrt{3-\sqrt5}\left(\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)^2+1\right)}\ dx-\frac{5+\sqrt5}{\sqrt{10(3+\sqrt5)}}\int_{0}^{\infty}\frac{\sqrt2}{\sqrt{3+\sqrt5}\left(\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)^2+1\right)}\ dx$$
$$$$Evaluating the indefinite integrals, we obtain:
$$\arctan\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)+C$$
$$$$Evaluating the improper integral:
$$\lim_{z \to \infty}\left[\arctan\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)\right]_0^z$$
$$=\lim_{z \to \infty}\left(\arctan\left(\frac{\sqrt2z}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2z}{\sqrt{3+\sqrt5}}\right)\right)$$
$$=\lim_{z \to \infty}\left(\arctan\left(\infty\right)-\arctan\left(\infty\right)\right)=\frac{\pi}2-\frac{\pi}2=0$$
The way that I evalauted the integral is pretty long and has a lot of calculations. Is there an easier way?
 A: Define $$f(x) = \frac{1-x^2}{1+3x^2+x^4}$$ so that $$f(x^{-1}) = -\frac{x^2(1-x^2)}{1+3x^2+x^4}.$$  Since $$\int_{x=0}^\infty f(x) \, dx = \int_{x=0}^1 f(x) \, dx + \int_{x=1}^\infty f(x) \, dx,$$ the transformation $$x = u^{-1}, \quad dx = -u^{-2} \, du$$ gives $$\int_{x=1}^\infty f(x) \, dx = \int_{u=1}^0  -\frac{u^2(1-u^2)}{1+3u^2+u^4} (-u^{-2}) \, du = -\int_{u=0}^1 f(u) \, du.$$  Thus, we immediately obtain cancellation and the integral of $f$ on $(0,\infty)$ is $0$.

For the indefinite integral, we write $$f(x) = \frac{x^{-2} - 1}{x^{-2} + 3 + x^2} = \frac{x^{-2} - 1}{1 + (x + x^{-1})^2}.$$  This suggests the substitution $$u = x + x^{-1}, \quad du = (1 - x^{-2}) \, dx,$$ giving $$\int f(x) \, dx = \int \frac{-du}{1+u^2} = -\tan^{-1} u + C = -\tan^{-1}\left(x + \frac{1}{x}\right) + C.$$
A: We have
\begin{align*}
\int \frac{1-x^2}{x^4 + 3x^2 + 1} \, dx
&= - \int \frac{1 - x^{-2}}{(x + x^{-1})^2 + 1} \, dx \\
&= - \arctan\left(x + \frac{1}{x}\right) + C.
\end{align*}
Here, we utilized the substitution $u = x + x^{-1}$ when we move to the second line. For the definite integral, we have
$$ \int_{0}^{\infty} \frac{1-x^2}{x^4 + 3x^2 + 1} \, dx = \left[ - \arctan\left(x + \frac{1}{x}\right) \right]_{0^+}^{+\infty} = -\frac{\pi}{2} + \frac{\pi}{2} = 0. $$
A: Substitute $x\mapsto\frac{1-x}{1+x}$ to reveal an integral of an odd function over an interval symmetric about the origin:
$$\begin{align*}
I &= \int_0^\infty \frac{1-x^2}{1+3x^2+x^4} \, dx \\[1ex]
&= 8 \int_{-1}^1 \frac x{4+6x^2+5x^4} \, dx \\[1ex]
&= \boxed{0}
\end{align*}$$
