How to prove that a function is in fact $L_1$? I'm working on two problems where I need to interchange the limit and integral signs, so I want to evoke Lebesgue's Dominated Convergence Theorem. I now know that the functions I have chosen do indeed dominate, but is there a way to show that the dominating functions are in fact Lebesgue integrable? (I don't want to calculate their integrals...)

One of the functions I am trying to show are Lebesgue integrable are: $\dfrac{1}
{1+x^2}$ over the domain $[0,\infty)$

How would I go about doing this?
Since you asked for more "theoretical" approach, maybe this is something for you.

We have 
\begin{equation}
\lim_{x \rightarrow \pm \infty}\frac{x^2}{1+x^2}=1
\end{equation}
and therefore
\begin{equation}
\frac{1}{1+x^2} \leq c \ \frac{1}{x^2},
\end{equation}
for $|x| \geq a$ and suitable constants $a$ and $c$.
On the (compact) interval $ [ -a,a ]$ we can bound the (continuous) function by a constant. And outside of the interval it is just $\frac{1}{x^2}$ scaled by a constant which is known to be integrable outside an interval around zero.

In this case it is quite easy to see what $c$ and $a$ are (even without the limit argument, as pointed out by David C. Ullrich) - but this approach may work in "more complicated" cases: The function considered is continuous and behaves like a integrable function when $x \rightarrow \pm \infty$.
If we're talking about $L^1(\mathbb{R})$, then we should show that the integral is finite (it is).

Also, recall that this neat function is also Riemann integrable - that means that it is Lebesgue integrable. Check this out.