RSA: must $d$ be an integer? I am only taking baby steps in RSA.

If $p=11$, $q=7$ and $e=3$, 

$$\phi(n) = 10*6 = 60$$

Then:

$$d = (2 (\phi(n)) + 1 ) / 3 = 121/3$$

Should $d$ be kept as a non-integer or is such a $d$ invalid?
$d$ must indeed be an integer.

To calculate $d$ you need to calculate $d=e^{-1}\bmod{\phi(n)}$ which is called the modular multiplicative inverse of $e\bmod{\phi(n)}$. For $d$ be computable you need to ensure that $$\gcd(e,\phi(n))=\gcd(e,(p-1)(q-1))=1$$

holds, which isn't the case with your sample parameters as $\gcd(3,60)=3\neq1$.

As fgrieu pointed out in the comments you can find the private exponent by finding the smallest $k$ for which $(k\cdot\phi(n)+1)/e$ is an integer. The run-time of this approach likely will grow linearly with the size of the $e$ value which is in the order of $2^{16}$ usually.

Of course $k$ isn't unique unless you take the smallest possible $k$, meaning there will be other (larger) values of $k$ yielding a valid decryption exponent.