# Prove that the identity is true ln(n+1)=Integral of f'(x)/f(x), if the limits of integration are x=n+1 and x=n+2 f(x) is a polynomial and it's roots are 1,2,3,....,n

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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.**