<s>[INST] Is this definite integral really independent of a parameter? How can it be shown? I want to find a nice simple expression for the definite integral
$$\int_0^\infty \frac{x^2\,dx}{(x^2-a^2)^2 + x^2}$$
Now, I can numerically compute this integral, and it seems to converge to $\pi/2$ for all real values of $a$. Is this integral actually always equal to $\pi/2$? How can I show this?
Also, why does Wolfram Alpha give me something that appears to depend on $a$? Is there a good reason it doesn't eliminate $a$?
 [/INST] Yes it is true! 
Let
$$\displaystyle I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$
Make the substitution $\displaystyle x = \dfrac{a^2}{t}$
We get
$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^6}{t^4\left(\dfrac{a^4}{t^2} + \left(\dfrac{a^4}{t^2} - a^2\right)^2\right)} \ \text{dt} = \int_{0}^{\infty} \dfrac{a^2}{t^2 + (a^2 - t^2)^2} \ \text{dt}$$
i.e.
$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2} \ \text{dx}$$
Therefore
$$\displaystyle 2I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx} + \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2}\ \text{dx}$$
$$ = \int_{0}^{\infty} \dfrac{x^2 + a^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$
$$\displaystyle = \int_{0}^{\infty} \dfrac{1 + \dfrac{a^2}{x^2}}{1 + \left(x-\dfrac{a^2}{x}\right)^2} \ \text{dx}$$
Making the substitution $\displaystyle t = x - \dfrac{a^2}{x}$
Gives us
$$\displaystyle 2I = \int_{-\infty}^{\infty} \dfrac{\text{dt}}{1 + t^2} = \pi$$
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