Question Number 112545 by mnjuly1970 last updated on 08/Sep/20
$$\:\:\:\:{please}\:{solve}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xlog}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx}\:=??? \\ $$$$ \\ $$$$\:\:\:\:\:\:…{m}.{n}.{july}\:\mathrm{1970}…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{good}\:\:\:{luck}\:. \\ $$$$ \\ $$
Commented by MJS_new last updated on 08/Sep/20
$$\mathrm{the}\:\mathrm{old}\:\mathrm{man}\:\mathrm{needs}\:\mathrm{no}\:\mathrm{luck},\:\mathrm{he}\:\mathrm{uses}\:\mathrm{his} \\ $$$$\mathrm{experience}!!! \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
$${peace}\:{be}\:{upon}\:{you}\:… \\ $$
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](https://www.tinkutara.com/question/Q112565.png)
$$\int{x}\mathrm{ln}^{\mathrm{2}} \:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{2}}{{x}+\mathrm{1}}\:\rightarrow\:{dx}=−\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}−\mathrm{2}}{{t}^{\mathrm{3}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:{dt}= \\ $$$$=\mathrm{2}\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt}−\mathrm{4}\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{3}} }{dt} \\ $$$$ \\ $$$$\mathrm{2}\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=−\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)}{dt}= \\ $$$$=−\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}−\mathrm{1}}{dt}−\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}}{dt} \\ $$$$ \\ $$$$−\mathrm{4}\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }−\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}{dt}= \\ $$$$=\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }−\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}−\mathrm{1}}{dt}+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}}{dt}+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{now}\:\mathrm{have} \\ $$$$\mathrm{2}\int\frac{{t}−\mathrm{2}}{{t}^{\mathrm{3}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:{dt}= \\ $$$$=−\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{2}\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt}= \\ $$$$=\mathrm{2}\frac{\mathrm{1}−{t}}{{t}^{\mathrm{2}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:+\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$ \\ $$$$\mathrm{4}\int\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=−\mathrm{4}\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{4}\int\frac{{dt}}{{t}\left({t}−\mathrm{1}\right)}= \\ $$$$=−\mathrm{4}\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{{t}}+\mathrm{4ln}\:\left({t}−\mathrm{1}\right)\:−\mathrm{4ln}\:{t} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{now}\:\mathrm{have} \\ $$$$\mathrm{2}\int\frac{{t}−\mathrm{2}}{{t}^{\mathrm{3}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:{dt}= \\ $$$$=\frac{\mathrm{4}\left({t}−\mathrm{1}\right)}{{t}}\mathrm{ln}\:\left({t}−\mathrm{1}\right)\:−\frac{\mathrm{2}\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} }\mathrm{ln}^{\mathrm{2}} \:\left({t}−\mathrm{1}\right)\:−\mathrm{4ln}\:{t}\:= \\ $$$$=\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:−\mathrm{2}\left({x}−\mathrm{1}\right)\mathrm{ln}\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:+\mathrm{4ln}\:\left({x}+\mathrm{1}\right)\:+{C} \\ $$$$ \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}\mathrm{ln}^{\mathrm{2}} \:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:{dx}=\mathrm{4ln}\:\mathrm{2} \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
$${hello}\:{sir} \\ $$$${im}\:{persian}\:. \\ $$$${im}\:{math}'{s}\:{teacher}. \\ $$$${in}\:{our}\:{language}\:'{ostad}' \\ $$$${means}\:{someone}\:{who} \\ $$$${knows}\:{a}\:{lot}\:{in}\:{her}\:/{his} \\ $$$${field}\:{and}\:{is}\:{a}\:{scholar}\:. \\ $$
Commented by Rasheed.Sindhi last updated on 08/Sep/20
$${Mahdi}\:{sir},\:{where}\:{are}\:{you}\:{from}? \\ $$$${Particularly}\:{I}\:{asked}\:{to}\:{see}\:{word} \\ $$$$“{ostad}'' \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
$${ahsant}\:\:{mercey}\:.{thank}\:{you} \\ $$$$\:{ostad}\:\left({master}\right) \\ $$
Commented by MJS_new last updated on 08/Sep/20
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}! \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
$${thank}\:\:{you}\:{so}\:{much} \\ $$$${again}.{you}\:{are}\:{a}\:{humble} \\ $$$${and}\:{noble}\:{man}\:. \\ $$
Commented by Rasheed.Sindhi last updated on 09/Sep/20
$$\mathcal{T}{hank}\:{you}\:{sir} \\ $$$$\mathcal{T}{he}\:{same}\:{word}\:{is}\:{used}\:{in}\:{my}\:{country} \\ $$$${in}\:{about}\:{same}\:{sense}\:{and}\:{also}\:{spoken}\:{for}\:'{teacher}'!{And}\:{many}\:{educated} \\ $$$${persons}\:{know}\:{that}\:{the}\:{word} \\ $$$${origially}\:{belongs}\:{to}\:{Persian}. \\ $$$$\mathcal{D}{ear}\:{sir}\:{I}\:{was}\:{also}\:{math}\:{teacher} \\ $$$${at}\:{the}\:{moment}\:{retired}. \\ $$$${I}\:{belong}\:{to}\:{your}\:{neighbouring} \\ $$$${country}\:{pakistan}. \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 09/Sep/20
$${ahsant}\:{bar}\:{shoma} \\ $$
Commented by Rasheed.Sindhi last updated on 09/Sep/20
$${Sir}\:{kaivan},\:{are}\:{you}\:{also} \\ $$$${persian}? \\ $$
Commented by mnjuly1970 last updated on 09/Sep/20
$${peace}\:{be}\:{upon}\:{you}\:\:{and}\:{all} \\ $$$${the}\:{honorable}\:{people}\:{of}\:{the} \\ $$$${great}\:{country}\:{and} \\ $$$${neighbour}\:{of} \\ $$$$\ast{PAKISTAN}\ast… \\ $$
Commented by kaivan.ahmadi last updated on 09/Sep/20
$${yes}\:{sir}. \\ $$
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 .](https://www.tinkutara.com/question/Q112588.png)
$$\mathrm{ln}\:\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={t} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}={e}^{{t}} \:\:\:\Rightarrow\:\:\:{x}=\frac{\mathrm{1}−{e}^{{t}} }{\mathrm{1}+{e}^{{t}} } \\ $$$${dx}=−\frac{\mathrm{2}{e}^{{t}} {dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{2}} } \\ $$$${question}:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\mathrm{ln}\:^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right){dx} \\ $$$$\Rightarrow\:\:{I}=\mathrm{2}\int\:\frac{{t}^{\mathrm{2}} \left({e}^{{t}} −\mathrm{1}\right){e}^{{t}} {dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{3}} }\:=\mathrm{2}\int{t}^{\mathrm{2}} {F}\left({t}\right){dt} \\ $$$$\int{F}\left({t}\right){dt}\:=\int\frac{\left({e}^{{t}} −\mathrm{1}\right){e}^{{t}} {dt}}{\left(\mathrm{1}+{e}^{{t}} \right)^{\mathrm{3}} }\:\:\:\:{let}\:\:{e}^{{t}} ={z} \\ $$$${t}=\mathrm{ln}\:{z}\:\:\:\Rightarrow\:\:\:{dt}=\frac{{dz}}{{z}} \\ $$$$\int{F}\left({t}\right){dt}\:=\int\:\left[\frac{\mathrm{1}}{\left({z}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}}{\left(\mathrm{1}+{z}\right)^{\mathrm{3}} }\right]{dz} \\ $$$$\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{1}+{z}}+\frac{\mathrm{1}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} }\:=−\frac{{z}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} } \\ $$$$\frac{{I}}{\mathrm{2}}={t}^{\mathrm{2}} \int{F}\left({t}\right){dt}−\mathrm{2}\int{t}\left\{\int{F}\left({t}\right){dt}\right\}{dt} \\ $$$$\:\:\:\:=\:\left(\mathrm{ln}\:{z}\right)^{\mathrm{2}} \left\{\frac{−{z}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} }\right\}\mid_{\mathrm{1}} ^{\mathrm{0}} +\mathrm{2}\int_{\mathrm{1}} ^{\:\mathrm{0}} \left(\mathrm{ln}\:{z}\right)\frac{{dz}}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} } \\ $$$$\:=\:\mathrm{0}+\mathrm{2}\left(\frac{\mathrm{ln}\:{z}}{\mathrm{1}+{z}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\int_{\mathrm{1}} ^{\:\mathrm{0}} \frac{{dz}}{{z}\left(\mathrm{1}+{z}\right)} \\ $$$$\:=\mathrm{0}+\left[\mathrm{2}\left(\frac{\mathrm{ln}\:{z}}{\mathrm{1}+{z}}\right)−\mathrm{2ln}\:{z}+\mathrm{2ln}\:\left(\mathrm{1}+{z}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:{I}\:=\:\mathrm{2}\left[−\frac{\mathrm{2}{z}\mathrm{ln}\:{z}}{\mathrm{1}+{z}}+\mathrm{2ln}\:\left(\mathrm{1}+{z}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:{I}\:=\mathrm{2}\left(\mathrm{0}+\mathrm{2ln}\:\mathrm{2}\right)\:=\:\mathrm{4ln}\:\mathrm{2}\:. \\ $$
Commented by mnjuly1970 last updated on 08/Sep/20
$${thank}\:{you}\: \\ $$$${excellent}\: \\ $$$${grateful}… \\ $$