Question Number 103196 by bobhans last updated on 13/Jul/20

∫_0 ^1  ((((1/2)−x) ln(1−x) dx)/(x^2 −x+1)) ?

Answered by bramlex last updated on 13/Jul/20

replace x with 1−x   I = ∫_0 ^1  (((x−(1/2)) ln x)/(x^2 −x+1)) dx [ by parts ]   { ((u = ln x)),((dv = (((x−(1/2)))/(x^2 −x+1)) dx)) :}  I = (1/2)ln (x) ln(x^2 −x+1)∣_0 ^1  −  (1/2)∫_0 ^1  ((ln(x^2 −x+1))/x) dx   I= −(1/2)∫_0 ^1 ((  ln(((x^3 +1)/(x+1))))/x) dx   I= (1/2)∫_0 ^1  ((ln(x+1))/x) dx  −(1/2)∫_0 ^1 ((ln(x^3 +1))/x) dx   I= (1/2)∫_0 ^1 (1/x)Σ_(n=1) ^∞ (((−1)^(n−1)  x^n )/n) dx −  (1/2)∫_0 ^1 (1/x)Σ_(n=1) ^∞ (((−1)^(n−1)  x^(3n) )/n)dx  I =(1/2)Σ_(n=1) ^∞ ∫_0 ^1 (((−1)^(n−1) x^(n−1) )/n) dx −  (1/2)Σ_(n=1) ^∞ ∫_0 ^1 (((−1)^(n−1) x^(3n−1) )/n) dx   I=(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )−  (1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/(3n))  I= ((1/2)−(1/6))Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )  I=(1/3)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) .  however Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =   Σ_(n=1) ^∞ (1/n^2 ) − Σ_(n=1) ^∞ (2/((2n)^2 ))  = (1−(1/2)) Σ_(n=1) ^∞ (1/n^2 ) = (1/2)Σ_(n=1) ^∞ (1/n^2 )  = (1/2)×(π^2 /6) = (π^2 /(12))  therefore we conclude that   I=∫_0 ^1  ((((1/2)−x)ln(1−x) dx)/(x^2 −x+1))  = (1/3)×(π^2 /(12)) = (π^2 /(36)) . ⊕

Commented byDwaipayan Shikari last updated on 13/Jul/20

Great  sir!