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∫_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 ) ..✓

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

Question Number 154208 Answers: 0 Comments: 0

∵∴∵∴prove that ∫_0 ^∞ ∫_0 ^∞ ((log(1−e^(−x) )(yLi_2 (e^(−x−y) )+Li_3 (e^(−x−y) ))/(1−e^(x+y) ))e^(x+y) dxdy=((21)/8)ζ(6)+ζ^2 (3) ∵∴∵by MATH.AMIN∴∵∴

$∵∴∵∴ p r o v e t h a t$ $\int_{0}^{\infty} \int_{0}^{\infty} \frac{l o g (1 - e^{- x}) ({y L i}_{2} (e^{- x - y}) + {L i}_{3} (e^{- x - y})}{1 - e^{x + y}} e^{x + y} d x d y = \frac{21}{8} ζ (6) + ζ^{2} (3)$ $∵∴∵ by MATH . AMIN ∴∵∴$

Question Number 154195 Answers: 3 Comments: 0

∫((x+sinx)/(1+cosx))dx please,help me

$\int \frac{x + s i n x}{1 + c o s x} d x$ $p l e a s e, h e l p m e$

Question Number 154192 Answers: 1 Comments: 0

Ω :=∫_0 ^( 1) (( x.sin(ln(x)))/(1−x))dx method 1 Ω= Im[∫_0 ^( 1) (( x^( i+1) )/(1−x)) dx=Φ] Φ = ∫_0 ^( 1) (( x^( i+1) +x^( i+2) )/(1−x^( 2) ))dx =^(x^( 2) =t) (1/2)∫_0 ^( 1) (( t^( (i/2)) −t^((i+1)/2) )/(1−t))dt = (1/2) { ψ (1 +((i+1)/2))−ψ (1+(i/2))} = (1/2) { (2/(1+i)) + ψ ( ((1+i)/2) )−(2/i)−ψ ((i/2))} = ((−i)/(−1+i)) + (1/2) {ψ (((1+i)/2))−ψ((i/2) )} = −(1/2) +(i/(2 )) +(1/2) {ψ(((1+i)/2))−ψ((i/2))} Ω = Im (Φ )= (1/2) +(1/2) Im(ψ((1/2) +(i/2)))−(1/2) Im((i/2)) = (1/(2 )) { 1 +(π/2) tanh((π/2)) −1−(π/2) coth((π/2))} =(π/4) {((−1)/(sinh((π/2)).cosh((π/2))))}=((−π)/(2 sinh (π ))) ✓

$Ω := \int_{0}^{1} \frac{x . s i n (l n (x))}{1 - x} d x$ $m e t h o d 1$ $Ω = I m [\int_{0}^{1} \frac{x^{i + 1}}{1 - x} d x = Φ]$ $Φ = \int_{0}^{1} \frac{x^{i + 1} + x^{i + 2}}{1 - x^{2}} d x$ $\overset{x^{2} = t}{=} \frac{1}{2} \int_{0}^{1} \frac{t^{\frac{i}{2}} - t^{\frac{i + 1}{2}}}{1 - t} d t$ $= \frac{1}{2} {ψ (1 + \frac{i + 1}{2}) - ψ (1 + \frac{i}{2})}$ $= \frac{1}{2} {\frac{2}{1 + i} + ψ (\frac{1 + i}{2}) - \frac{2}{i} - ψ (\frac{i}{2})}$ $= \frac{- i}{- 1 + i} + \frac{1}{2} {ψ (\frac{1 + i}{2}) - ψ (\frac{i}{2})}$ $= - \frac{1}{2} + \frac{i}{2} + \frac{1}{2} {ψ (\frac{1 + i}{2}) - ψ (\frac{i}{2})}$ $Ω = I m (Φ) = \frac{1}{2} + \frac{1}{2} I m (ψ (\frac{1}{2} + \frac{i}{2})) - \frac{1}{2} I m (\frac{i}{2})$ $= \frac{1}{2} {1 + \frac{π}{2} t a n h (\frac{π}{2}) - 1 - \frac{π}{2} c o t h (\frac{π}{2})}$ $= \frac{π}{4} {\frac{- 1}{s i n h (\frac{π}{2}) . c o s h (\frac{π}{2})}} = \frac{- π}{2 s i n h (π)} ✓$