Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.

Sure, here is a Python solution using recursion and backtracking:

def permute(nums):
    def backtrack(first = 0):
        # if all integers are used up
        if first == n:  
            output.append(nums[:])
        for i in range(first, n):
            # place i-th integer first 
            # in the current permutation
            nums[first], nums[i] = nums[i], nums[first]
            # use next integers to complete the permutations
            backtrack(first + 1)
            # backtrack
            nums[first], nums[i] = nums[i], nums[first]
            
    n = len(nums)
    output = []
    backtrack()
    return output

Here's how it works:

1. If all integers are used up, add the permutation to the output.
2. Iterate over the integers from first to the end of array.
3. Place the integer at index first that we're iterating on at the beginning of the permutation, i.e. swap(nums[first], nums[i]).
4. Proceed to create all permutations which starts from first and keeps first+1 in the remaining part.
5. Now we have to backtrack, so we swap the numbers back to restore the original array. We do this because we want to try a different possibility for the number at the first index.
6. Repeat the process until we have found all permutations.