Set up the limit \(\displaystyle{ \lim_{x \to \infty} { \frac{f(x)}{g(x)} } }\). Use the hint for \( g(x) = 1/\ln x\).

\(\displaystyle{ L = \lim_{x \to \infty}{ \frac{1/(2+\cos x + \ln x)}{1/\ln x} } }\)

\(\displaystyle{ L = \lim_{x \to \infty}{ \frac{1}{(2+\cos x + \ln x)/(\ln x)} } }\)

\(\displaystyle{ L = \lim_{x \to \infty}{ \frac{1}{(2/\ln x)+(\cos x/\ln x) + 1} } }\)

As \(x\) gets very large, \(\displaystyle{ \lim_{x \to \infty}{ 2/\ln x } = 0 }\)

And, as \(x\) gets very large, \(\displaystyle{ \lim_{x \to \infty}{ \cos x/\ln x } = 0 }\)

This leaves \(L = 1\).

Since \(L\) is finite and non-zero, then the original integral will converge or diverge the same as \(\displaystyle{ \int_3^{\infty}{g(x)~dx} }\). So now we need to determine if this integral converges or diverges.

We can directly compare \(y=\ln x\) with \(y=x\). Since \(\ln x \lt x\), \( 1/x \lt 1/\ln x\).

Also, \(\displaystyle{ \int_3^{\infty}{ 1/x ~dx } \lt \int_3^{\infty}{ 1/\ln x ~ dx } }\)

Since \(\displaystyle{ \int_3^{\infty}{ 1/x ~dx } }\) diverges, then \(\displaystyle{\int_3^{\infty}{ 1/\ln x ~ dx } }\) also diverges and so does the original integral.

We can see that the integrand is positive on the integration interval. So now we need to find a comparison function \(g(x)\). As we look at the integrand, the \(x^5\) under the radical dominates the \(x^3\) term as \(x\) gets very large. So we can ignore the \(x^3\) term. This leaves \(x/x^{5/2} = 1/x^{3/2}\). So we will use \(g(x) = 1/x^{3/2}\) as our comparison function.

Now let's set up the limit \(\displaystyle{ L = \lim_{x\to\infty}{f(x)/g(x)} }\)

\(\displaystyle{ L = \lim_{x\to\infty}{ (x/\sqrt{x^5+x^3})/(1/x^{3/2}) } }\)

\(\displaystyle{ L = \lim_{x\to\infty}{ \frac{x^{5/2}}{\sqrt{x^5+x^3}} } }\)

\(\displaystyle{ L = \lim_{x\to\infty}{ \frac{1}{\sqrt{1+1/x}} } = 1 }\)

The limit is real, finite and positive. So now we need to determine if the integral of \(g(x)\) converges or diverges.

Fortunately, we can evaluate \(\displaystyle{ \int_3^{\infty}{ 1/x^{3/2} ~dx } }\)

\(\displaystyle{ \int_3^{\infty}{ x^{-3/2} ~dx } }\)

\(\displaystyle{ \lim_{b\to\infty}{\int_3^{b}{ x^{-3/2} ~dx }} }\)

\(\displaystyle{ \lim_{b\to\infty}\left[ \frac{x^{-1/2}}{-1/2} \right]_3^{b} }\)

\(\displaystyle{ \lim_{b\to\infty}\left[ \frac{-2}{x^{1/2}} \right]_3^{b} }\)

\(\displaystyle{ \lim_{b\to\infty}\frac{-2}{(b)^{1/2}} - \frac{-2}{3^{1/2}} }\)

\( 0 + 2/\sqrt{3} = 2/\sqrt{3} \)

Since the comparison integral converged, so does our original integral.