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Question Number 163144 by HongKing last updated on 04/Jan/22

$Find : Ω = \underset{0}{\int^{1}} \frac{\frac{π}{4} - \arctan (x)}{1 - x} dx$
	Answered by mnjuly1970 last updated on 04/Jan/22
	$−−− solution Ω= ∫_0 ^( 1) (( artan(((1−x)/(1+x))))/(1−x)) dx ((1−x)/(1+x )) = t ⇒ dx= ((−2)/((1+t )^( 2) ))dt Ω = 2∫_0 ^( 1) (( (arctan( t ))/(2t(1+t ))) dt =2 ∫_0 ^( 1) {(( arctan(t ))/t) −((arctan(t))/(1+t)) }dt Φ= ∫_0 ^( 1) (( arctan(t))/t) dt , Ψ =∫_0 ^( 1) ((arctan(t))/(1+t))dt Φ= ∫_0 ^( 1) Σ_(n=0) ^∞ (( (−1)^( n) t^( 2n) )/(2n+1))dt = G=β(2) Ψ = [ ln(1+t).arctan(t)]_0 ^1 −∫_0 ^( 1) ((ln(1+ t))/(1+t^( 2) )) = (π/8) ln(2) Ω= 2G − (π/4) ln(2) ■ m.n$
	$- - -$ $s o l u t i o n$ $Ω = \int_{0}^{1} \frac{a r t a n (\frac{1 - x}{1 + x})}{1 - x} d x$ $\frac{1 - x}{1 + x} = t \Rightarrow d x = \frac{- 2}{{(1 + t)}^{2}} d t$ $Ω = 2 \int_{0}^{1} \frac{(a r c t a n (t)}{2 t (1 + t)} d t$ $= 2 \int_{0}^{1} {\frac{a r c t a n (t)}{t} - \frac{a r c t a n (t)}{1 + t}} d t$ $Φ = \int_{0}^{1} \frac{a r c t a n (t)}{t} d t, Ψ = \int_{0}^{1} \frac{a r c t a n (t)}{1 + t} d t$ $Φ = \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} t^{2 n}}{2 n + 1} d t = G = β (2)$ $Ψ = {[l n (1 + t) . a r c t a n (t)]}_{0}^{1} - \int_{0}^{1} \frac{l n (1 + t)}{1 + t^{2}}$ $= \frac{π}{8} l n (2)$ $Ω = 2 G - \frac{π}{4} l n (2) ◼ m . n$
	Commented by HongKing last updated on 04/Jan/22

	$perfect solution my dear Sir, thank you so much$
	Commented by mnjuly1970 last updated on 04/Jan/22

	$h a v e a n i c e t i m e s i r$
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