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Question Number 109838 by mnjuly1970 last updated on 25/Aug/20
please prove::: ∫_0 ^( 1) (1/( (√(1−x))))log((x/(1−x)))dx =4log(2)
pleaseprove:::∫0111−xlog(x1−x)dx=4log(2)
Answered by mathmax by abdo last updated on 25/Aug/20
I =∫_0 ^1 (1/( (√(1−x))))ln((x/(1−x)))dx changement (√(1−x))=t give 1−x =t^2 ⇒ I =−∫_0 ^1 (1/t)ln(((1−t^2 )/t^2 ))(−2t)dt =2 ∫_0 ^1 (ln(1−t^2 )−2ln(t))dt =2∫_0 ^1 ln(1−t^2 )dt −4∫_0 ^1 ln(t)dt we have ∫_0 ^1 ln(t)dt =[tlnt−t]_0 ^1 =−1 ∫_0 ^1 ln(1−t^2 )dt =_(by pafts) [(t−1) ln(1−t^2 )]_0 ^1 −∫_0 ^1 (t−1)×((−2t)/(1−t^2 ))dt =−2 ∫_0 ^1 ((t(1−t))/(1−t^2 ))dt =−2∫_0 ^1 (t/(1+t)) dt =−2∫_0 ^1 ((1+t−1)/(1+t)) dt =−2 +2ln(2) ⇒I =−4 +4ln(2)−4(−1) =4ln(2) ⇒ ★ ∫_0 ^1 (1/( (√(1−x))))ln((x/(1−x)))dx =4ln(2)★
I=∫0111−xln(x1−x)dxchangement1−x=tgive1−x=t2⇒I=−∫011tln(1−t2t2)(−2t)dt=2∫01(ln(1−t2)−2ln(t))dt=2∫01ln(1−t2)dt−4∫01ln(t)dtwehave∫01ln(t)dt=[tlnt−t]01=−1∫01ln(1−t2)dt=bypafts[(t−1)ln(1−t2)]01−∫01(t−1)×−2t1−t2dt=−2∫01t(1−t)1−t2dt=−2∫01t1+tdt=−2∫011+t−11+tdt=−2+2ln(2)⇒I=−4+4ln(2)−4(−1)=4ln(2)⇒★∫0111−xln(x1−x)dx=4ln(2)★
Commented by mnjuly1970 last updated on 26/Aug/20
thank you so much . excellent^∞
thankyousomuch.excellent∞
Commented by mathmax by abdo last updated on 27/Aug/20
you are welcome
youarewelcome

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