Generally, finding a ring of integers is rather annoying to do by hand. Basically you compute relative discriminants (or equivalently norms of differents) and localize (and complete at the primes ofc) to compute the discriminants separately according to whether it's tamely ramified, wildly ramified, unramified, or totally ramified.
The last step is generally easy, except when it's wildly ramified, for example if you have totally ramified of deg p over \mathbb{Q}_p, p prime. But I can leave this as an exercise to show that for galois case, totally ramified deg p gives valuation of the discriminant to be 2p-2 (there are two descriptions of discriminant, one in terms of norm of f'(a) for integral generator a, and another in terms of the roots of f(a)).
But for quadratic case, you can do this by hand and the ring of integers in \mathbb{Q}(\sqrt{2}) is just \mathbb{Z}[\sqrt{2}].
So for example, try to compute the ring of integers in \mathbb{Q}(17^{1/3}). The discriminant of Z[17^(1/3)] is 3^3 * 17^2, and 3 is not totally ramified when you localize and complete at (3), so Z[17^(1/3)] can't be all of the ring of integers. Now you manually look for more elements, and find that (1 + 17^(1/3))^2 / 3 is integral over Z, adjoin it, and note that this time the discriminant is 3*17^2, so that's the ring of integers.
