Since the polynomial has n roots, we'll write the polynomial
f(x) as a product of linear factors:
f(x) =
a(x-1)(x-2)...(x-n)
We'll note each factor as a
function:
x-1 = f1 => f1' =
1
x-2 = f2 => f2' =
1
.............
x-n = fn => fn'
= 1
f(x) = f1*f2*...*fn
We'll
differentiate both sides, using product rule:
f'(x) =
(f1*f2*...*fn)'
(f1*f2*...*fn)' = f1'*f2*..fn + f1*f2'*f3*....fn +
... + f1*f2*...*fn'
f'(x) = (x-2)*...*(x-n) + (x-1)*(x-3)*...*(x-n)
+ ... + (x-1)*...(x-n+1)
We'll divide by f(x) both
sides:
f'(x)/f(x) = 1/(x-1) + 1/(x-2) + ... +
1/(x-n)
We'll integrate both sides:
Int
f'(x)dx/f(x) =Int dx/(x-1) + Int dx/(x-2) + ... + Int dx/(x-n)
Int
f'(x)dx/f(x) =ln(x-1) + ln(x-2) + ... + ln(x-n)
We'll apply Leibniz
Newton:
Int f'(x)dx/f(x) =ln(n+1)/n + ln(n)/(n-1) + ... +
ln2/1
Int f'(x)dx/f(x) =
ln[(n+1)n(n-1).....2/n(n-1)....1]
Int f'(x)dx/f(x) =
ln[(n+1)!/n!]
But (n+1)! =n!*(n+1) => ln[(n+1)!/n!] =
ln[n!*(n+1)/n!]
Int f'(x)dx/f(x) =
ln(n+1)
The identity ln(n+1)=Integral of f'(x)/f(x) is
true, for the polynomial function f(x), whose roots are
1,2,...,n.
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