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Lol no, am happy to study in Australia. The exams are still nice for practice though.

x is an integer, and a,b,c,d distinct integers, so x-a,x-b,x-c,x-d are distinct integer factors of 4, with product 4. A quick consideration of cases tells you these factors must be 2,-2,1,-1. Then sum the 4 equations to get 4x-(a+b+c+d)=2-2+1-1=0, and you are done. If you are looking for some...

Yeah, it is nothing special. And "academic journal" is a stretch for a mathematics education journal lol.

Yep, works super well for these integrals with logs.

I'll leave the intended solution for someone else to write out (p.s I think you need to be a bit careful about the first part, the integrals are indefinite so equality will only hold up to a constant as written.) A cute alternate solution (but I don't believe this technique is allowed in MX2):

It is just tedious... Edit: Sorry, typoed the numerator of the second integral in the second line of the evaluation of I, should just be cos(2x).