The question:

Use the inequality form of the comparison test to determine whether or not the following improper integral converges

$\displaystyle \int_{1}^{\infty} \frac{1}{\sqrt{1 + x^4}} dx$

My attempt:

When x is large, dominant term is $\displaystyle x^4$

$\displaystyle \frac{1}{\sqrt{1 + x^4}} < \frac{1}{\sqrt{x^4}}$

$\displaystyle \int_{1}^{\infty} \frac{1}{x^2} dx$

$\displaystyle _{1}^{\infty}[\frac{-1}{x}]$

As x -> infinity, function goes to 0:

0 - (-1)

So the integral converges (approaches 1). Is this correct? Thanks.