Show how to calculate definite integral for x=1 or x=2 of function f(x) if f(x)=8x^2+4-3f(1/x)?

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This was an interesting one. It took me a while to see the trick. I tried to integrate too soon.

Since

f(x)=8x^2+4x-3f(1/x) this means

f(1/x) = 8/x^2 + 4/x - 3f(x) evaluating the above at 1/x

This means

f(x) = 8x^2 + 4x - 3(8/x^2 + 4/x - 3f(x))

So

f(x) = 8x^2 + 4x - 24/x^2 - 12/x + 9f(x)

we get

-8f(x) = 8x^2 + 4x - 24/x^2 - 12/x

So

f(x) = -x^2 - 1/2x + 3/x^2 + 3/(2x)

`int^2_1 f(x) dx = int^2_1 (-x^2 - 1/2x + 3/x^2 + 3/(2x)) dx`

`= (-1/3x^3 - 1/4x^2 - 1/3(3/x^3) + 3/2 ln(x))|^2_1`

`=-1/3(8)-1/4(4)-1/3(3/8)+3/2ln(2)+1/3(1)+1/4(1)+1/3(3/1)+3/2ln(1)=3/4ln2-19/12`

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