Given, (7^1/2 + 5^1/3 )³⁷

Now, general term of this binomial T_r+1 = ³⁷C_r * (7^1/2 )^37-r  * (5^1/3 )^r

=> T_r+1 = ³⁷C_r * 7^(37-r)/2  * (5)^r/3

This General term will be an integer if ³⁷C_r is an integer, 7^(37-r)/2  is an integer and (5)^r/3 is an integer.

Now, ³⁷C_r will always be a positive integer.

Since ³⁷C_r denotes the number of ways of selecting r things out of 37 things, it can not be a fraction.

So, ³⁷C_r is an integer.

Again, 7^(37-r)/2  will be an integer if (37 - r)/2 is an integer.

So, r = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37    ...................1

And if (5)^r/3 is an integer, then r/3 shold be an integer.

So, r = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36     .............2

Now, take intersection of 1 and 2, we get

r = 3, 9, 15, 21, 27, 33

So, total possible value of r is 6

Hence, there are 6 integers are in the binomial expansion of (7^1/2 + 5^1/3 )³⁷