Let d = X2 + 1 
R[X].
X + (d)
R[X]/(d) is invertible
in R[X]/(d).
X3
R[X]/(d) is invertible.
X
Q[X]/(X2)
is not invertible.
We have
X · (-X) = -X2 =
1.
So the inverse of
X + (d) is -X.
Using the extended Euclidean algorithm we find the equality
X · X3 + (1 - X2)(X2 + 1)
= 1. Going over to residue classes modulo
X2 + 1, we find
X · X3 = 1
in R[X]/(d).
So an inverse of
X3 is X.
Of course we can also first find a representative of
X3 of degree smaller than 2 and work with that.
Indeed, there is no polynomial
a such that Xa - 1 is divisible by
X2. In other words, there is no a
satisfying X · a = 1.