Improper integrals and comparison tests

Question

I have this problem:  Evaluate the integral (x^2-2)/(x^4+3) from 1 to infinity using a comparison test (evaluate if converging).  I understand the basic idea is to find a function with a graph "above/below" this function that can be integrated.  My solutions guide uses (3x^2)/(x^4) which of course is just 3x^-2.  I see that it's easy to find the antiderivative of this and take it's limit at infinity.  How did they come up with 3x^-2 as a comparison?  Am I just missing some simple Algebra?

Answer 1

Notice that the numerator is positive for all x > √2 and the denominator is positive for all real x.

In a fraction with both numerator and denominator being positive, increasing the numerator or decreasing the denominator will increase the value.

e.g. 2/3 > 2/5 because 3 < 5 and 3/5 > 2/5 because 3 > 2.

So using this when x > √2:

We can increase the numerator by 2 to x², and decrease the denominator by 3 to x⁴.

We wind up with x²/ x⁴ > (x²-2) / (x⁴+3).

So we can use x²/x⁴ = x^-2 as an upper bound for this interval. (3x^-2 is usable too since it is even bigger than x^-2).

The integrand is negative for 1<x<√2 so our bound applies in that region too and therefore is an upper bound for all x>1.

Thus ∫₁^∞ (x²-2)/(x⁴+3) dx < ∫₁^∞ x^-2 dx = [-x^-1]₁^∞ = 0 - (-1) = 1, which proves convergence.