This is not valid C++11 code, because a constexpr function must only contain a return statement.
This is incorrect. static_assert in a constexpr function are fine. What is not fine is using function parameters in constant expressions, like you do it.
You could throw if x <= 0. Calling the function in a context that requires a constant expression will then fail to compile
constexpr int do_something(int x) {
return x > 0 ? (x + 5) : (throw std::logic_error("x must be > 0"));
}