Ω = ∫_0 ^( 1) (( x.tanh^( −1) (x))/((1+x)^( 2) ))dx= (1/(24)) (π^( 2) −6)


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Question Number 176542 by mnjuly1970 last updated on 20/Sep/22

$Ω = \int_{0}^{1} \frac{x . {t a n h}^{- 1} (x)}{{(1 + x)}^{2}} d x = \frac{1}{24} (π^{2} - 6)$
	Answered by Peace last updated on 20/Sep/22
	$∫_0 ^1 (x/((1+x)^2 ))tg^− (x)dx tanh^− (x)=(1/2)(ln(1+x)−ln(1−x)) (x/((1+x)^2 ))=(1/(1+x))−(1/((1+x)^2 )) 2Ω=∫_0 ^1 ((ln(1+x))/(1+x))−((ln(1+x))/((1+x)^2 ))dx+∫_0 ^1 ((ln(1−x))/((1+x)^2 ))dx−∫_0 ^1 ((ln(1−x))/(1+x))dx =(1/2)[ln^2 (1+x)]_0 ^1 +[((ln(1+x))/(1+x))]_0 ^1 −∫_0 ^1 (1/((1+x)^2 ))+A+B =((ln^2 (2))/2)+((ln(2))/2)+(1/2)+A+B A=lim_(x→1) ∫_0 ^x ((ln(1−t))/((1+t)^2 ))dt=lim_(x→1) {−((ln(1−x))/(1+x))−∫_0 ^x (1/(1−t^2 )).dt} =lim_(x→1) −((ln(1−x))/(1+x))−(1/2)∫_0 ^x (1/(1−t))+(1/(1+t))dt =lim_(x→1) −((ln(1−x))/(1+x))+((ln(1−x))/2)−((ln(1+x))/2) =−((ln(2))/2) −B=∫_0 ^1 ((ln(1−x))/(1+x))dx x=((1−t)/(1+t)),=−1+(2/(1+t))⇒dt=−((2dt)/((1+t)^2 )) 2∫_0 ^1 ((ln(((2t)/(1+t))))/(2/((1+t)))).(1/((1+t)^2 ))dt=∫_0 ^1 ((ln(2t)−ln(1+t))/(1+t))dt =∫_0 ^1 ((ln(2))/(1+t))+((ln(t))/(1+t))−((ln(1+t))/(1+t))dt =ln^2 (2)−((ln^2 (2))/2)+∫_0 ^1 ((ln(t))/(1+t))dt=((ln^2 (2))/2)+∫_0 ^1 ((ln(1−(−t))/(−t))d(−t) =((ln^2 (2))/2)−Li_2 (−1) B=−((ln^2 (2))/2)+Li_2 (−1) 2Ω=((ln^2 (2))/2)+((ln(2))/2)+(1/2)+A+B =((ln^2 (2))/2)+((ln(2))/2)+(1/2)−((ln(2))/2)−((ln^2 (2))/2)+Li_2 (−1) =(1/2)+Li_2 (−1)=(1/2)+(π^2 /(12))=(1/(12))(π^2 −6) Ω=(1/(24))(π^2 −6)$
	$\int_{0}^{1} \frac{x}{{(1 + x)}^{2}} {t g}^{-} (x) d x$ ${t a n h}^{-} (x) = \frac{1}{2} (l n (1 + x) - l n (1 - x))$ $\frac{x}{{(1 + x)}^{2}} = \frac{1}{1 + x} - \frac{1}{{(1 + x)}^{2}}$ $2 Ω = \int_{0}^{1} \frac{l n (1 + x)}{1 + x} - \frac{l n (1 + x)}{{(1 + x)}^{2}} d x + \int_{0}^{1} \frac{l n (1 - x)}{{(1 + x)}^{2}} d x - \int_{0}^{1} \frac{l n (1 - x)}{1 + x} d x$ $= \frac{1}{2} {[{l n}^{2} (1 + x)]}_{0}^{1} + {[\frac{l n (1 + x)}{1 + x}]}_{0}^{1} - \int_{0}^{1} \frac{1}{{(1 + x)}^{2}} + A + B$ $= \frac{{l n}^{2} (2)}{2} + \frac{l n (2)}{2} + \frac{1}{2} + A + B$ $A = \lim_{x \to 1} \int_{0}^{x} \frac{l n (1 - t)}{{(1 + t)}^{2}} d t = \lim_{x \to 1} {- \frac{l n (1 - x)}{1 + x} - \int_{0}^{x} \frac{1}{1 - t^{2}} . d t}$ $= \lim_{x \to 1} - \frac{l n (1 - x)}{1 + x} - \frac{1}{2} \int_{0}^{x} \frac{1}{1 - t} + \frac{1}{1 + t} d t$ $= \lim_{x \to 1} - \frac{l n (1 - x)}{1 + x} + \frac{l n (1 - x)}{2} - \frac{l n (1 + x)}{2}$ $= - \frac{l n (2)}{2}$ $- B = \int_{0}^{1} \frac{l n (1 - x)}{1 + x} d x$ $x = \frac{1 - t}{1 + t}, = - 1 + \frac{2}{1 + t} \Rightarrow d t = - \frac{2 d t}{{(1 + t)}^{2}}$ $2 \int_{0}^{1} \frac{l n (\frac{2 t}{1 + t})}{\frac{2}{(1 + t)}} . \frac{1}{{(1 + t)}^{2}} d t = \int_{0}^{1} \frac{l n (2 t) - l n (1 + t)}{1 + t} d t$ $= \int_{0}^{1} \frac{l n (2)}{1 + t} + \frac{l n (t)}{1 + t} - \frac{l n (1 + t)}{1 + t} d t$ $= {l n}^{2} (2) - \frac{{l n}^{2} (2)}{2} + \int_{0}^{1} \frac{l n (t)}{1 + t} d t = \frac{{l n}^{2} (2)}{2} + \int_{0}^{1} \frac{l n (1 - (- t)}{- t} d (- t)$ $= \frac{{l n}^{2} (2)}{2} - {L i}_{2} (- 1)$ $B = - \frac{{l n}^{2} (2)}{2} + {L i}_{2} (- 1)$ $2 Ω = \frac{{l n}^{2} (2)}{2} + \frac{l n (2)}{2} + \frac{1}{2} + A + B$ $= \frac{{l n}^{2} (2)}{2} + \frac{l n (2)}{2} + \frac{1}{2} - \frac{l n (2)}{2} - \frac{{l n}^{2} (2)}{2} + {L i}_{2} (- 1)$ $= \frac{1}{2} + {L i}_{2} (- 1) = \frac{1}{2} + \frac{π^{2}}{12} = \frac{1}{12} (π^{2} - 6)$ $Ω = \frac{1}{24} (π^{2} - 6)$
	Commented by Tawa11 last updated on 22/Sep/22

	$Great sir$
	Commented by Peace last updated on 23/Sep/22

	$t h a n k Y o u n i c e D a y$
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