can someone show me the proof of this mathematical term ( A = X^1-y / 1-y ) if y=1 then A = ln(X)

What you wrote looks off.  I'm guessing you mean

lim y→1 (x^(1-y)-1)/(1-y) = ln(x)

That's the limit definition for the derivative of x^(y) with respect to y at y = 1. It follows from x^(y) = e^(ln{x}y) and derivative of e^(y) and chain rule, all calculus topics.
The power rule for integration states ∫x^(n)dx = x^(n+1)/(n+1) + C if and only if n ≠ -1, as x^(0)/0 is undefined. You’re wondering why ∫x^(-1)dx = ln(x) + C. It’s actually a definition but a proof requires knowing some intro to calculus:

> y = ln(x) means x = e^(y). Then d(x)/dx = e^(y) * dy/dx which implies dy/dx = e^(-y) = e^(-ln&#40;x&#41;) = 1/x. Then by the fundamental theorem of calculus, ∫x^(-1)dx = 1/x + C.
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