Is it ok to set $g(x)=1$ for the quotient test of convergence of improper integrals?
I find it easily solves many problems, for example, show if the following converge or diverge:
$\displaystyle\int^{\infty}_1\frac {\arctan x}{x\ln ^2(1+x)}dx$
So, $g(x)=1$, $\displaystyle\lim_{x\to\infty} \frac {f(x)}{g(x)}=\lim_{x\to\infty} \frac {\arctan x}{x\ln ^2(1+x)}= 0$ ($\arctan x$ is bounded and the denominator tends to zero).
Since $\displaystyle\int^{\infty}_1 1dx$ converges so does the given.
The integral $\int_1^\infty 1\,dx$ does not converge. so the procedure will not work in this case.
A good $g(x)$ here would be $\frac{1}{(1+x)\ln^2(1+x)}$, whose integral converges by a direct calculation.