∫_0 ^( 1) (x^( i+1) /(1−x)) dx = [−ln(1−x)x^( i+1) ]_0 ^( 1) + (1+i)∫_0 ^( 1) x^( i) ln (1−x )dx = (1+i ) ∫_0 ^( 1) Σ_(n=1) ^∞ (x^( n+i) /n)dx = (1+i )Σ(1/(n (n+i+1 )))= = Σ(1/(n )) −(1/(n+i+1)) = γ + ψ ( i+2 ) = γ + (1/(1+i)) +ψ (1+i) γ + ((1−i)/2) + (1/i) + ψ (i ) im = −(3/2) + (1/2) +(π/2) coth(π ) −1 +(π/2) tan h(π ) ..✓


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Question Number 154223 by mnjuly1970 last updated on 15/Sep/21

$\int_{0}^{1} \frac{x^{i + 1}}{1 - x} d x = {[- l n (1 - x) x^{i + 1}]}_{0}^{1}$ $+ (1 + i) \int_{0}^{1} x^{i} l n (1 - x) d x$ $= (1 + i) \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{x^{n + i}}{n} d x$ $= (1 + i) Σ \frac{1}{n (n + i + 1)} =$ $= Σ \frac{1}{n} - \frac{1}{n + i + 1} = γ + ψ (i + 2)$ $= γ + \frac{1}{1 + i} + ψ (1 + i)$ $γ + \frac{1 - i}{2} + \frac{1}{i} + ψ (i)$ $i m = - \frac{3}{2} + \frac{1}{2} + \frac{π}{2} c o t h (π)$ $- 1 + \frac{π}{2} t a n h (π) . . ✓$
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