Let x represent the smaller integer. Then the other one is 4x+1 (1 more than 4 times the first one). Their product will be ...
... x(4x+1) = 39 . . . . . . an algebraic equation to solve
_____
In standard form, this is ...
... 4x² +x -39 = 0
... (x -3)(4x +13) = 0 . . . . . factored form
... x = 3 or x = -13/4
The only integer solution here is x = 3. Then the other is 4·3+1 = 13.
The two integers are 3 and 13.
_____
Comment on factoring
There are a couple of different ways you can factor the equation. They basically involve looking for factors of 4×39 that differ by 1. Of course 39 = 3×13, so you are looking for factors of 4×3×13 = 12×13 that are different by 1. Guess what? You've found them! (12 and 13).
One method of factoring requires you use these to rewrite the equation as ...
... 4x² -12x +13x -39 = 0
then factor by grouping.
... 4x(x -3) +13(x -3) = 0 = (4x+13)(x -3)
___
Another method of factoring lets you use the leading coefficient in both factors, then divide by that coefficient in some convenient way:
... (4x-12)(4x+13)/4 = 0 = (x -3)(4x +13) . . . . . note the use of 4x in both factors