Test the following improper integral for convergence:
![\[ \int_0^{\infty} \frac{x}{\sqrt{x^4+1}} \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8f5367053608da2b8dedc187b177ee7d_l3.png "Rendered by QuickLaTeX.com")
This integral diverges.
Proof. By example 1 (page 417) of Apostol, we know the integral 
diverges. We then apply the limit comparison test (Theorem 10.25 on page 418),
![\begin{align*} \lim_{x \to +\infty} \frac{f(x)}{g(x)} &= \lim_{x \to +\infty} \frac{\frac{x}{\sqrt{x^4+1}}}{\frac{1}{x}} \\[9pt] &= \lim_{x \to +\infty} \frac{x^2}{\sqrt{x^4+1}} \\[9pt] &= \lim_{x \to +\infty} \frac{1}{\sqrt{1+\frac{1}{x^4}}} \\[9pt] &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-63723874cb32c1bad4dc9feb9cb00512_l3.png "Rendered by QuickLaTeX.com")
Thus, by the limit comparison test, we have established the divergence of
![\[ \int_0^{\infty} \frac{x}{\sqrt{x^4+1}} \, dx. \qquad \blacksquare \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-01d3124e6d89fcca74748f6d73431ace_l3.png "Rendered by QuickLaTeX.com")