<s>[INST] Integral of a rational function: Proof of $\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$? I suspect that
$$\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}=
 {{\pi}\over{2}}$$
for $C>0$.
I tried $C=1$, $C=2$, $C=42$, and $C=\frac{1}{1000}$ with Wolfram Alpha. But how to prove it?
 [/INST] Let $f : \Bbb{R} \to \Bbb{C}$ be an integrable even function. Then we claim that
$$ \int_{0}^{\infty} f\left( y - \frac{1}{y} \right) \, dy = \int_{0}^{\infty} f(x) \, dx. \tag{1} $$
Indeed, let $I$ denote the LHS of $(1)$. Then by the substitution $y \mapsto y^{-1}$ we have
$$ I = \int_{0}^{\infty} f\left( \frac{1}{y} - y \right) \, \frac{dy}{y^2} = \int_{0}^{\infty} \frac{1}{y^2} f \left( y - \frac{1}{y} \right) \, dy. $$
Thus with the substitution $x = y - y^{-1}$, we have
$$ 2I = \int_{0}^{\infty} \left( 1 + \frac{1}{y^2} \right) f\left( y - \frac{1}{y} \right) \, dy = \int_{-\infty}^{\infty} f(x) \, dx = 2 \int_{0}^{\infty} f(x) \, dx $$
and the conclusion follows.
Now you can plug
$$ f(x) = \frac{1}{C + x^2} $$
to $(1)$ to obtain the conclusion:
$$ \int_{0}^{\infty} \frac{y^2}{Cy^2 + (y^2 - 1)^2} \, dy = \int_{0}^{\infty} \frac{dy}{C + (y - y^{-1})^2} \, dy = \int_{0}^{\infty} \frac{dx}{C + x^2} = \frac{\pi}{2\sqrt{C}}. $$
</s>