Are integrable functions always bounded?

Yes, an integrable function can be unbounded. For example, the function $1/\sqrt{x}$ on the domain (0,1] is unbounded but the integral has a finite value.

I suppose that you are talking about the Riemann integral here. If so, yes, the concept is defined only for bounded functions defined on intervals which are closed and bounded, that is, intervals of the type $[a,b]$, with $a<b$.

If $f$ is unbounded or if the interval is unbounded, we get the so-called improper integrals. For instance, we sey that $\frac1{x^2}$ is integrable on $[1,+\infty)$, because the limit$$\lim_{M\to\infty}\int_1^M\frac{\mathrm dx}{x^2}$$exists and we define$$\int_1^{+\infty}\frac{\mathrm dx}{x^2}=\lim_{M\to\infty}\int_1^M\frac{\mathrm dx}{x^2}=1.$$But this is an extension of the concept of Riemann integral.