Question: I have simple integral that must be solved by substitution.
$$\int \frac{1}{\sqrt{1-2x-x^2}} \, dx = \int \frac{1}{\sqrt{2-(x+1)^2}} \, dx $$
After substitution $$u = x+1 $$
we get
$$\int \frac{1}{\sqrt{-u^2+2}} \, du $$
But i dont know how to continue. Thanks for any help.

Answer: 
Indeed the following result holds:

**Theorem**:
A Hamming-Code ${\mathrm Ham} [n,n-\ell]\_q$ is equivalent to a cyclic code 
if and only if $gcd(\ell,q-1)=1$.

For $n=4$, $q=3$ and $\ell=2$ we obtain, for example, that the ternary Hamming code ${\mathrm Ham} [4,2]\_3$ is not equivalent to a cyclic code, since $gcd(2,2)=2\neq 1$.
The smallest counterexample is the Hamming code of codimension $2$ over $\mathbb{F}\_3$, as written in the answer of Dietrich Burde.
This code has the parameters $[4,2,3]\_3$.

To see that this Hamming code cannot be brought into cyclic form, we check the ternary cyclic $[4,2]\_3$-codes, which are the candidates. From the factorization $X^4 - 1 = (X-1)(X+1)(X^2 + 1)$ into monic irreducible factors over $\mathbb{F}\_3$, we see that there are two cyclic $[4,2]\_3$ codes, given by the generator polynomials $(X^2 + 1)$ and $(X-1)(X+1) = X^2 - 1$. However, both generator polynomials have the Hamming weight $2$, so the generated codes cannot have minimum distance $3$.