Read This First: Here are my first thoughts. The biggest problem with this is that it's not rigorous and utilizes a lot of 'hand-waving arguments' (and I'm not positive that this hand-waving is even justified here!). Another problem with it is that it's not really all that simple, so even if it can be justified, I don't think that it's much good as a solution. I think that hamster143 outlined a much better solution too. But with that qualified, here was my first thought about the problem.
Let the function f be defined such that
f(x) = \frac{a(x^2-1)^2-2x(x+a)^2}{(x+a)^3(ax+1)^3}
Since f is continuous on [0,\infty), we can apply the second fundamental theorem of calculus to find that ...
\int_0^{\infty}f(x)\mathrm{d}x = \lim_{x \to \infty}F(t) - F(0)
where F is an anti-derivative of f. Therefore, we need only find the values of these anti-derivatives. We'll go about this in an indirect way.
To find F(0), first define the function g such that g(x) = a(x^2-1)^2-2x(x+a)^2. Next, note that by choosing x small enough, we can find numbers h,k > 0 such that the following inequality holds:
h[g(x)] \leq f(x) \leq k[g(x)]
Since it's easy to verify that G(0) = 0 (neglecting the constant) where G is an anti-derivative of g, this suggests that the anti-derivative of f at zero is equal to zero. Therefore, F(0) = 0.
To evaluate the term \lim_{x \to \infty}F(t), we can note that as x becomes arbitrarily large, f(x) tends to something like h(x) = x^{-2} (because the numerator is a polynomial of degree 4 and the denominator is a polynomial of degree 6). Since lim_{x \to \infty}H(x) = 0 (once again, neglecting the constant) where H is an anti-derivative of h, this suggests that \lim_{x \to \infty}F(t) = 0.
Combining these two results, we find that ...
\int_0^{\infty}f(x)\mathrm{d}x = 0
You make it sound like there's One True Antiderivative which all others are just modifications of!
