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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)

$$ \\ $$$$ \\ $$$$\:\:\:\:{please}\:{prove}::: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}}}{log}\left(\frac{{x}}{\mathrm{1}−{x}}\right){dx}\:=\mathrm{4}{log}\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$

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)★

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{x}}}\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}}\right)\mathrm{dx}\:\:\:\mathrm{changement}\:\sqrt{\mathrm{1}−\mathrm{x}}=\mathrm{t}\:\mathrm{give}\:\mathrm{1}−\mathrm{x}\:=\mathrm{t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{I}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{t}}\mathrm{ln}\left(\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} }\right)\left(−\mathrm{2t}\right)\mathrm{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{ln}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)−\mathrm{2ln}\left(\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{ln}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}\:−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}\:\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}\:=\left[\mathrm{tlnt}−\mathrm{t}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}\:=_{\mathrm{by}\:\mathrm{pafts}} \:\:\:\:\left[\left(\mathrm{t}−\mathrm{1}\right)\:\mathrm{ln}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{t}−\mathrm{1}\right)×\frac{−\mathrm{2t}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}\left(\mathrm{1}−\mathrm{t}\right)}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}}\:\mathrm{dt}\:=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\mathrm{t}−\mathrm{1}}{\mathrm{1}+\mathrm{t}}\:\mathrm{dt} \\ $$$$=−\mathrm{2}\:+\mathrm{2ln}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{I}\:=−\mathrm{4}\:+\mathrm{4ln}\left(\mathrm{2}\right)−\mathrm{4}\left(−\mathrm{1}\right)\:=\mathrm{4ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\bigstar\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{x}}}\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}}\right)\mathrm{dx}\:=\mathrm{4ln}\left(\mathrm{2}\right)\bigstar \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 26/Aug/20

thank you so much .  excellent^∞

$${thank}\:{you}\:{so}\:{much}\:. \\ $$$${excellent}^{\infty} \\ $$

Commented by mathmax by abdo last updated on 27/Aug/20

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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