<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] So, you can start out by writing the denominator like this: $ y^4+y^2(C-2)+1=(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})$. This is simply the solution of a quadratic equation (ugly, but it will do). So, now, we want to simplify this expression somehow, like by writing:
$ \frac{y^2}{(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}=\frac{A}{y^2-\frac{2-C-\sqrt{C(C-4)}}{2}}+\frac{B}{y^2-\frac{2-C+\sqrt{C(C-4)}}{2}}=\frac{A(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})+B(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})}{(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}=\frac{y^2(A+B)-1(A(\frac{2-C-\sqrt{C(C-4)}}{2})+B(\frac{2-C-\sqrt{C(C-4)}}{2})}{(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})}$. 
Therefore,we want $A+B=1 \wedge A(\frac{2-C+\sqrt{C(C-4)}}{2})+B(\frac{2-C-\sqrt{C(C-4}}{2})=0$. This is a linear system, that you can solve to find A and B, and then calculate the integral. 
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