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Question Number 112545 by mnjuly1970 last updated on 08/Sep/20
    please solve :         I=∫_0 ^( 1) xlog^2 (((1−x)/(1+x)))dx =???          ...m.n.july 1970....             good   luck .
$$\:\:\:\:{please}\:{solve}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xlog}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx}\:=??? \\ $$$$ \\ $$$$\:\:\:\:\:\:…{m}.{n}.{july}\:\mathrm{1970}…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{good}\:\:\:{luck}\:. \\ $$$$ \\ $$
Commented by MJS_new last updated on 08/Sep/20
the old man needs no luck, he uses his  experience!!!
$$\mathrm{the}\:\mathrm{old}\:\mathrm{man}\:\mathrm{needs}\:\mathrm{no}\:\mathrm{luck},\:\mathrm{he}\:\mathrm{uses}\:\mathrm{his} \\ $$$$\mathrm{experience}!!! \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
peace be upon you ...
$${peace}\:{be}\:{upon}\:{you}\:… \\ $$
Answered by MJS_new last updated on 08/Sep/20
∫xln^2  ((1−x)/(1+x)) dx=       [t=(2/(x+1)) → dx=−(((x+1)^2 )/2)dt]  =2∫((t−2)/t^3 )ln^2  (t−1) dt=  =2∫((ln^2  (t−1))/t^2 )dt−4∫((ln^2  (t−1))/t^3 )dt    2∫((ln^2  (t−1))/t^2 )dt=       [by parts]  =−2((ln^2  (t−1))/t)+4∫((ln (t−1))/(t(t−1)))dt=  =−2((ln^2  (t−1))/t)+4∫((ln (t−1))/(t−1))dt−4∫((ln (t−1))/t)dt    −4∫((ln^2  (t−1))/t^3 )dt=       [by parts]  =2((ln^2  (t−1))/t^2 )−4∫((ln (t−1))/(t^2 (t−1)))dt=  =2((ln^2  (t−1))/t^2 )−4∫((ln (t−1))/(t−1))dt+4∫((ln (t−1))/t)dt+4∫((ln (t−1))/t^2 )dt    ⇒ we now have  2∫((t−2)/t^3 )ln^2  (t−1) dt=  =−2((ln^2  (t−1))/t)+2((ln^2  (t−1))/t^2 )+4∫((ln (t−1))/t^2 )dt=  =2((1−t)/t^2 )ln^2  (t−1) +4∫((ln (t−1))/t^2 )dt    4∫((ln (t−1))/t^2 )dt=       [by parts]  =−4((ln (t−1))/t)+4∫(dt/(t(t−1)))=  =−4((ln (t−1))/t)+4ln (t−1) −4ln t    ⇒ we now have  2∫((t−2)/t^3 )ln^2  (t−1) dt=  =((4(t−1))/t)ln (t−1) −((2(t−1))/t^2 )ln^2  (t−1) −4ln t =  =(((x^2 −1))/2)ln^2  ((1−x)/(1+x)) −2(x−1)ln ((1−x)/(1+x)) +4ln (x+1) +C    ⇒ ∫_0 ^1 xln^2  ((1−x)/(1+x)) dx=4ln 2
$$\int{x}\mathrm{ln}^{\mathrm{2}} \:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{2}}{{x}+\mathrm{1}}\:\rightarrow\:{dx}=−\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}−\mathrm{2}}{{t}^{\mathrm{3}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:{dt}= \\ $$$$=\mathrm{2}\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt}−\mathrm{4}\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{3}} }{dt} \\ $$$$ \\ $$$$\mathrm{2}\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=−\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt}= \\ $$$$=−\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}−\mathrm{1}}{dt}−\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}}{dt} \\ $$$$ \\ $$$$−\mathrm{4}\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }−\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}{dt}= \\ $$$$=\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }−\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}−\mathrm{1}}{dt}+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}}{dt}+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{now}\:\mathrm{have} \\ $$$$\mathrm{2}\int\frac{{t}−\mathrm{2}}{{t}^{\mathrm{3}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:{dt}= \\ $$$$=−\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt}= \\ $$$$=\mathrm{2}\frac{\mathrm{1}−{t}}{{t}^{\mathrm{2}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$ \\ $$$$\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=−\mathrm{4}\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{4}\int\frac{{dt}}{{t}\left({t}−\mathrm{1}\right)}= \\ $$$$=−\mathrm{4}\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{4ln}\:\left({t}−\mathrm{1}\right)\:−\mathrm{4ln}\:{t} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{now}\:\mathrm{have} \\ $$$$\mathrm{2}\int\frac{{t}−\mathrm{2}}{{t}^{\mathrm{3}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:{dt}= \\ $$$$=\frac{\mathrm{4}\left({t}−\mathrm{1}\right)}{{t}}\mathrm{ln}\:\left({t}−\mathrm{1}\right)\:−\frac{\mathrm{2}\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:−\mathrm{4ln}\:{t}\:= \\ $$$$=\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:−\mathrm{2}\left({x}−\mathrm{1}\right)\mathrm{ln}\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:+\mathrm{4ln}\:\left({x}+\mathrm{1}\right)\:+{C} \\ $$$$ \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}\mathrm{ln}^{\mathrm{2}} \:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:{dx}=\mathrm{4ln}\:\mathrm{2} \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
hello sir  im persian .  im math′s teacher.  in our language ′ostad′  means someone who  knows a lot in her /his  field and is a scholar .
$${hello}\:{sir} \\ $$$${im}\:{persian}\:. \\ $$$${im}\:{math}'{s}\:{teacher}. \\ $$$${in}\:{our}\:{language}\:'{ostad}' \\ $$$${means}\:{someone}\:{who} \\ $$$${knows}\:{a}\:{lot}\:{in}\:{her}\:/{his} \\ $$$${field}\:{and}\:{is}\:{a}\:{scholar}\:. \\ $$
Commented by Rasheed.Sindhi last updated on 08/Sep/20
Mahdi sir, where are you from?  Particularly I asked to see word  “ostad”
$${Mahdi}\:{sir},\:{where}\:{are}\:{you}\:{from}? \\ $$$${Particularly}\:{I}\:{asked}\:{to}\:{see}\:{word} \\ $$$$“{ostad}'' \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
ahsant  mercey .thank you   ostad (master)
$${ahsant}\:\:{mercey}\:.{thank}\:{you} \\ $$$$\:{ostad}\:\left({master}\right) \\ $$
Commented by MJS_new last updated on 08/Sep/20
you′re welcome!
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}! \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
thank  you so much  again.you are a humble  and noble man .
$${thank}\:\:{you}\:{so}\:{much} \\ $$$${again}.{you}\:{are}\:{a}\:{humble} \\ $$$${and}\:{noble}\:{man}\:. \\ $$
Commented by Rasheed.Sindhi last updated on 09/Sep/20
Thank you sir  The same word is used in my country  in about same sense and also spoken for ′teacher′!And many educated  persons know that the word  origially belongs to Persian.  Dear sir I was also math teacher  at the moment retired.  I belong to your neighbouring  country pakistan.
$$\mathcal{T}{hank}\:{you}\:{sir} \\ $$$$\mathcal{T}{he}\:{same}\:{word}\:{is}\:{used}\:{in}\:{my}\:{country} \\ $$$${in}\:{about}\:{same}\:{sense}\:{and}\:{also}\:{spoken}\:{for}\:'{teacher}'!{And}\:{many}\:{educated} \\ $$$${persons}\:{know}\:{that}\:{the}\:{word} \\ $$$${origially}\:{belongs}\:{to}\:{Persian}. \\ $$$$\mathcal{D}{ear}\:{sir}\:{I}\:{was}\:{also}\:{math}\:{teacher} \\ $$$${at}\:{the}\:{moment}\:{retired}. \\ $$$${I}\:{belong}\:{to}\:{your}\:{neighbouring} \\ $$$${country}\:{pakistan}. \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 09/Sep/20
ahsant bar shoma
$${ahsant}\:{bar}\:{shoma} \\ $$
Commented by Rasheed.Sindhi last updated on 09/Sep/20
Sir kaivan, are you also  persian?
$${Sir}\:{kaivan},\:{are}\:{you}\:{also} \\ $$$${persian}? \\ $$
Commented by mnjuly1970 last updated on 09/Sep/20
peace be upon you  and all  the honorable people of the  great country and  neighbour of  ∗PAKISTAN∗...
$${peace}\:{be}\:{upon}\:{you}\:\:{and}\:{all} \\ $$$${the}\:{honorable}\:{people}\:{of}\:{the} \\ $$$${great}\:{country}\:{and} \\ $$$${neighbour}\:{of} \\ $$$$\ast{PAKISTAN}\ast… \\ $$
Commented by kaivan.ahmadi last updated on 09/Sep/20
yes sir.
$${yes}\:{sir}. \\ $$
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 .
$$\mathrm{ln}\:\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={t} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}={e}^{{t}} \:\:\:\Rightarrow\:\:\:{x}=\frac{\mathrm{1}−{e}^{{t}} }{\mathrm{1}+{e}^{{t}} } \\ $$$${dx}=−\frac{\mathrm{2}{e}^{{t}} {dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{2}} } \\ $$$${question}:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\mathrm{ln}\:^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx} \\ $$$$\Rightarrow\:\:{I}=\mathrm{2}\int\:\frac{{t}^{\mathrm{2}} \left({e}^{{t}} −\mathrm{1}\right){e}^{{t}} {dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{3}} }\:=\mathrm{2}\int{t}^{\mathrm{2}} {F}\left({t}\right){dt} \\ $$$$\int{F}\left({t}\right){dt}\:=\int\frac{\left({e}^{{t}} −\mathrm{1}\right){e}^{{t}} {dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{3}} }\:\:\:\:{let}\:\:{e}^{{t}} ={z} \\ $$$${t}=\mathrm{ln}\:{z}\:\:\:\Rightarrow\:\:\:{dt}=\frac{{dz}}{{z}} \\ $$$$\int{F}\left({t}\right){dt}\:=\int\:\left[\frac{\mathrm{1}}{\left({z}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{\left(\mathrm{1}+{z}\right)^{\mathrm{3}} }\right]{dz} \\ $$$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{1}+{z}}+\frac{\mathrm{1}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} }\:=−\frac{{z}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} } \\ $$$$\frac{{I}}{\mathrm{2}}={t}^{\mathrm{2}} \int{F}\left({t}\right){dt}−\mathrm{2}\int{t}\left\{\int{F}\left({t}\right){dt}\right\}{dt} \\ $$$$\:\:\:\:=\:\left(\mathrm{ln}\:{z}\right)^{\mathrm{2}} \left\{\frac{−{z}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} }\right\}\mid_{\mathrm{1}} ^{\mathrm{0}} +\mathrm{2}\int_{\mathrm{1}} ^{\:\mathrm{0}} \left(\mathrm{ln}\:{z}\right)\frac{{dz}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} } \\ $$$$\:=\:\mathrm{0}+\mathrm{2}\left(\frac{\mathrm{ln}\:{z}}{\mathrm{1}+{z}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\int_{\mathrm{1}} ^{\:\mathrm{0}} \frac{{dz}}{{z}\left(\mathrm{1}+{z}\right)} \\ $$$$\:=\mathrm{0}+\left[\mathrm{2}\left(\frac{\mathrm{ln}\:{z}}{\mathrm{1}+{z}}\right)−\mathrm{2ln}\:{z}+\mathrm{2ln}\:\left(\mathrm{1}+{z}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:{I}\:=\:\mathrm{2}\left[−\frac{\mathrm{2}{z}\mathrm{ln}\:{z}}{\mathrm{1}+{z}}+\mathrm{2ln}\:\left(\mathrm{1}+{z}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:{I}\:=\mathrm{2}\left(\mathrm{0}+\mathrm{2ln}\:\mathrm{2}\right)\:=\:\mathrm{4ln}\:\mathrm{2}\:. \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
thank you   excellent   grateful...
$${thank}\:{you}\: \\ $$$${excellent}\: \\ $$$${grateful}… \\ $$

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