It helps, in this case, to ask what exactly a rational number is, and how they come up. The various number systems we work with in math come up in response to different kinds of algebraic problems. The natural numbers (sometimes abbreviated N) are the most basic, and they include all of the "counting numbers" you learn in early gradeschool - 1, 2, 3, 4, etc. - and occasionally 0. If we add or multiply any two natural numbers together, we get another natural number, so we don't need to worry about creating any new numbers when we're just working with those operations. Subtraction poses a few issues, though. If we take an expression like 3 - 4, our answer is going to be outside of N. Rather than throwing our hands up and calling it a day at this point, we create a new set of numbers to deal with this situation.
The integers (abbreviated Z) contain all of the natural numbers, and add negative numbers, too (-1, -2, -3, etc.). Multiplication, addition, and subtraction work fine now, and no matter which integers we use, those operations will always give us back another integer. It's not all perfect, though; division gives us a bit of a hassle now. Expressions like 6/3 are fine, since they just produce another integer (2, in this case), but there's no integer result for something like 2/5 or 3/8, so again, we have to create a new set of numbers to compensate.
That set of numbers is what we refer to as the rational numbers (abbreviated Q), named after the fact that all of them are expressed as ratios. By expanding our number system with these, we gain the ability to express every integer as a rational number by putting it over 1 (6 = 6/1, 5 = 5/1, etc.).
So, getting back to the original question, "is the quotient of an integer x and 13 a rational number?" Yes. The quotient of two integers can always be represented as a rational number.
