please solve : I=∫_0 ^( 1) xlog^2 (((1−x)/(1+x)))dx =??? ...m.n.july 1970.... good luck .


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Question Number 112545 by mnjuly1970 last updated on 08/Sep/20

$p l e a s e s o l v e :$ $I = \int_{0}^{1} {x l o g}^{2} (\frac{1 - x}{1 + x}) d x = ? ? ?$ $. . . m . n . j u l y 1970 . . . .$ $g o o d l u c k .$
Commented by MJS_new last updated on 08/Sep/20

$the old man needs no luck, he uses his$ $experience!!!$
Commented by mnjuly1970 last updated on 08/Sep/20

$p e a c e b e u p o n y o u . . .$
	Answered by MJS_new last updated on 08/Sep/20

	$\int x \ln^{2} \frac{1 - x}{1 + x} d x =$ $[t = \frac{2}{x + 1} \to d x = - \frac{{(x + 1)}^{2}}{2} d t]$ $= 2 \int \frac{t - 2}{t^{3}} \ln^{2} (t - 1) d t =$ $= 2 \int \frac{\ln^{2} (t - 1)}{t^{2}} d t - 4 \int \frac{\ln^{2} (t - 1)}{t^{3}} d t$ $2 \int \frac{\ln^{2} (t - 1)}{t^{2}} d t =$ $[by parts]$ $= - 2 \frac{\ln^{2} (t - 1)}{t} + 4 \int \frac{\ln (t - 1)}{t (t - 1)} d t =$ $= - 2 \frac{\ln^{2} (t - 1)}{t} + 4 \int \frac{\ln (t - 1)}{t - 1} d t - 4 \int \frac{\ln (t - 1)}{t} d t$ $- 4 \int \frac{\ln^{2} (t - 1)}{t^{3}} d t =$ $[by parts]$ $= 2 \frac{\ln^{2} (t - 1)}{t^{2}} - 4 \int \frac{\ln (t - 1)}{t^{2} (t - 1)} d t =$ $= 2 \frac{\ln^{2} (t - 1)}{t^{2}} - 4 \int \frac{\ln (t - 1)}{t - 1} d t + 4 \int \frac{\ln (t - 1)}{t} d t + 4 \int \frac{\ln (t - 1)}{t^{2}} d t$ $\Rightarrow we now have$ $2 \int \frac{t - 2}{t^{3}} \ln^{2} (t - 1) d t =$ $= - 2 \frac{\ln^{2} (t - 1)}{t} + 2 \frac{\ln^{2} (t - 1)}{t^{2}} + 4 \int \frac{\ln (t - 1)}{t^{2}} d t =$ $= 2 \frac{1 - t}{t^{2}} \ln^{2} (t - 1) + 4 \int \frac{\ln (t - 1)}{t^{2}} d t$ $4 \int \frac{\ln (t - 1)}{t^{2}} d t =$ $[by parts]$ $= - 4 \frac{\ln (t - 1)}{t} + 4 \int \frac{d t}{t (t - 1)} =$ $= - 4 \frac{\ln (t - 1)}{t} + 4 \ln (t - 1) - 4 \ln t$ $\Rightarrow we now have$ $2 \int \frac{t - 2}{t^{3}} \ln^{2} (t - 1) d t =$ $= \frac{4 (t - 1)}{t} \ln (t - 1) - \frac{2 (t - 1)}{t^{2}} \ln^{2} (t - 1) - 4 \ln t =$ $= \frac{(x^{2} - 1)}{2} \ln^{2} \frac{1 - x}{1 + x} - 2 (x - 1) \ln \frac{1 - x}{1 + x} + 4 \ln (x + 1) + C$ $\Rightarrow \underset{0}{\int^{1}} x \ln^{2} \frac{1 - x}{1 + x} d x = 4 \ln 2$
	Commented by mnjuly1970 last updated on 08/Sep/20

	$h e l l o s i r$ $i m p e r s i a n .$ $i m {m a t h}^{'} s t e a c h e r .$ $i n o u r l a n g u a g e^{'} {o s t a d}^{'}$ $m e a n s s o m e o n e w h o$ $k n o w s a l o t i n h e r / h i s$ $f i e l d a n d i s a s c h o l a r .$
	Commented by Rasheed.Sindhi last updated on 08/Sep/20

	$M a h d i s i r, w h e r e a r e y o u f r o m ?$ $P a r t i c u l a r l y I a s k e d t o s e e w o r d$ $‘ ‘ {o s t a d}^{″}$
	Commented by mnjuly1970 last updated on 08/Sep/20

	$a h s a n t m e r c e y . t h a n k y o u$ $o s t a d (m a s t e r)$
	Commented by MJS_new last updated on 08/Sep/20

	${you}^{'} re welcome!$
	Commented by mnjuly1970 last updated on 08/Sep/20

	$t h a n k y o u s o m u c h$ $a g a i n . y o u a r e a h u m b l e$ $a n d n o b l e m a n .$
	Commented by Rasheed.Sindhi last updated on 09/Sep/20

	$T h a n k y o u s i r$ $T h e s a m e w o r d i s u s e d i n m y c o u n t r y$ $i n a b o u t s a m e s e n s e a n d a l s o s p o k e n f o r^{'} {t e a c h e r}^{'}! A n d m a n y e d u c a t e d$ $p e r s o n s k n o w t h a t t h e w o r d$ $o r i g i a l l y b e l o n g s t o P e r s i a n .$ $D e a r s i r I w a s a l s o m a t h t e a c h e r$ $a t t h e m o m e n t r e t i r e d .$ $I b e l o n g t o y o u r n e i g h b o u r i n g$ $c o u n t r y p a k i s t a n .$
	Commented by kaivan.ahmadi last updated on 09/Sep/20

	$a h s a n t b a r s h o m a$
	Commented by Rasheed.Sindhi last updated on 09/Sep/20

	$S i r k a i v a n, a r e y o u a l s o$ $p e r s i a n ?$
	Commented by mnjuly1970 last updated on 09/Sep/20

	$p e a c e b e u p o n y o u a n d a l l$ $t h e h o n o r a b l e p e o p l e o f t h e$ $g r e a t c o u n t r y a n d$ $n e i g h b o u r o f$ $* P A K I S T A N * . . .$
	Commented by kaivan.ahmadi last updated on 09/Sep/20

	$y e s s i r .$
	Answered by ajfour last updated on 08/Sep/20
	$ln (((1−x)/(1+x)))=t ⇒ ((1−x)/(1+x))=e^t ⇒ x=((1−e^t )/(1+e^t )) dx=−((2e^t dt)/((1+e^t )^2 )) question: I=∫_0 ^( 1) xln^2 (((1−x)/(1+x)))dx ⇒ I=2∫ ((t^2 (e^t −1)e^t dt)/((1+e^t )^3 )) =2∫t^2 F(t)dt ∫F(t)dt =∫(((e^t −1)e^t dt)/((1+e^t )^3 )) let e^t =z t=ln z ⇒ dt=(dz/z) ∫F(t)dt =∫ [(1/((z+1)^2 ))−(2/((1+z)^3 ))]dz = −(1/(1+z))+(1/((1+z)^2 )) =−(z/((1+z)^2 )) (I/2)=t^2 ∫F(t)dt−2∫t{∫F(t)dt}dt = (ln z)^2 {((−z)/((1+z)^2 ))}∣_1 ^0 +2∫_1 ^( 0) (ln z)(dz/((1+z)^2 )) = 0+2(((ln z)/(1+z)))∣_0 ^1 +2∫_1 ^( 0) (dz/(z(1+z))) =0+[2(((ln z)/(1+z)))−2ln z+2ln (1+z)]_0 ^1 I = 2[−((2zln z)/(1+z))+2ln (1+z)]_0 ^1 I =2(0+2ln 2) = 4ln 2 .$
	$\ln (\frac{1 - x}{1 + x}) = t$ $\Rightarrow \frac{1 - x}{1 + x} = e^{t} \Rightarrow x = \frac{1 - e^{t}}{1 + e^{t}}$ $d x = - \frac{2 e^{t} d t}{{(1 + e^{t})}^{2}}$ $q u e s t i o n : I = \int_{0}^{1} x \ln^{2} (\frac{1 - x}{1 + x}) d x$ $\Rightarrow I = 2 \int \frac{t^{2} (e^{t} - 1) e^{t} d t}{{(1 + e^{t})}^{3}} = 2 \int t^{2} F (t) d t$ $\int F (t) d t = \int \frac{(e^{t} - 1) e^{t} d t}{{(1 + e^{t})}^{3}} l e t e^{t} = z$ $t = \ln z \Rightarrow d t = \frac{d z}{z}$ $\int F (t) d t = \int [\frac{1}{{(z + 1)}^{2}} - \frac{2}{{(1 + z)}^{3}}] d z$ $= - \frac{1}{1 + z} + \frac{1}{{(1 + z)}^{2}} = - \frac{z}{{(1 + z)}^{2}}$ $\frac{I}{2} = t^{2} \int F (t) d t - 2 \int t {\int F (t) d t} d t$ $= {(\ln z)}^{2} {\frac{- z}{{(1 + z)}^{2}}} ∣_{1}^{0} + 2 \int_{1}^{0} (\ln z) \frac{d z}{{(1 + z)}^{2}}$ $= 0 + 2 (\frac{\ln z}{1 + z}) ∣_{0}^{1} + 2 \int_{1}^{0} \frac{d z}{z (1 + z)}$ $= 0 + {[2 (\frac{\ln z}{1 + z}) - 2 \ln z + 2 \ln (1 + z)]}_{0}^{1}$ $I = 2 {[- \frac{2 z \ln z}{1 + z} + 2 \ln (1 + z)]}_{0}^{1}$ $I = 2 (0 + 2 \ln 2) = 4 \ln 2 .$
	Commented by mnjuly1970 last updated on 08/Sep/20

	$t h a n k y o u$ $e x c e l l e n t$ $g r a t e f u l . . .$
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