Using std::enable_if with anonymous type parameters

I try to use std::enable_if with an unused and unnamed type parameter, in order to not distort the return type. However, the following code does not compile.

#include <iostream>

template <typename T, typename = std::enable_if_t<!std::is_integral<T>::value>>
T foo() { std::cout << "non-integral" << std::endl; return T(); }

template <typename T, typename = std::enable_if_t<std::is_integral<T>::value>>
T foo() { std::cout << "integral" << std::endl; return T(); }

int main() {
  foo<float>();
  foo<int>();
}

The compiler says:

7:3: error: redefinition of 'template<class T, class> T foo()'
4:3: note: 'template<class T, class> T foo()' previously declared here
 In function 'int main()':
11:12: error: no matching function for call to 'foo()'
11:12: note: candidate is:
4:3: note: template<class T, class> T foo()
4:3: note: template argument deduction/substitution failed:

What is the problem here? How do I have to change the code to get it compile? The text book "Discovering Modern C++" explicitly encourages the use of std::enable_if with anonymous type parameters.