Let one integer be represented by n. The other integer must be 4n+1 from the description.
The product is 39 so n(4n+1)=39, and expanding the bracket we get 4n²+n=39. This can be written 4n²+n-39=0. This should factorise, but we can use a trick: 39=3×13 so we can expect the answer to involve at least one of these numbers. 4n²+n-39=0=(n-3)(4n+13). Sure enough, n-3 is a factor so n-3=0 is a solution and that means n=3 is one integer. The other is 4×3+1=13 (surprise, surprise!) and the integers are indeed 3 and 13.