0-1-x-1-x-ln-x-1-x-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 97552 by bemath last updated on 08/Jun/20 ∫10x1−xln(x1−x)dx? Answered by john santu last updated on 08/Jun/20 Answered by abdomathmax last updated on 08/Jun/20 letI=∫01x1−xln(x1−x)dxchangementx1−x=tgivex1−x=t2⇒x=t2−t2x⇒(1+t2)x=t2⇒x=t21+t2=1+t2−11+t2=1−11+t2⇒dxdt=2t(1+t2)2⇒I=∫0∞tln(t21+t21−t21+t2)2t(1+t2)2dt=4∫0∞t2(1+t2)2ln(t)dt=4∫0∞1+t2−1(1+t2)2ln(t)dt=4∫0∞lnt1+t2dt−4∫0∞lnt(1+t2)2dt=0−4∫0∞lnt(1+t2)2dtletA=∫0∞lnt(1+t2)2dt⇒A=∫01lnt(1+t2)2dt+∫1+∞lnt(1+t2)2dt∫1+∞lnt(1+t2)2dt=t=1u∫01lnu(1+1u2)2×−duu2=−∫01u2lnu(1+u2)2du⇒A=∫01(1−t2)lnt(1+t2)2dt11+u=∑n=0∞(−1)nun⇒−1(1+u)2=∑n=1∞n(−1)nun−1⇒1(1+u)2=∑n=1∞n(−1)n−1un−1⇒1(1+x2)2=∑n=1∞n(−1)n−1x2n−2⇒A=∫01(1−x2)lnx(∑n=1∞n(−1)n−1x2n−2)dx=∑n=1∞n(−1)n−1∫01(x2n−2−x2n)lnxdxwn=∫01(x2n−2−x2n)ln(x)dx=[(12n−1x2n−1−12n+1x2n+1)lnx]01−∫01(12n−1x2n−2−12n+1x2n)dx=−1(2n−1)2+1(2n+1)2⇒A=∑n=1∞n(−1)n−1(1(2n+1)2−1(2n−1)2)=∑n=1∞n(2n+1)2(−1)n−1−∑n=1∞n(−1)n−1(2n−1)2(→n=p+1)=∑n=1∞n(2n+1)2(−1)n−1−∑p=0∞(p+1)(−1)p(2p+1)2=−∑n=1∞n(−1)n(2n+1)2−∑n=1∞(n+1)(−1)n(2n+1)2−1=−2∑n=1∞n(−1)n(2n+1)2−∑n=1∞(−1)n(2n+1)2−1resttofindthosesums…becontinued… Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-97550Next Next post: Find-all-the-triples-of-positive-integers-x-y-z-so-that-x-y-2020-y-z-2020-is-a-rational-number-and-x-2-y-2-z-2-be-a-prime-number- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![let I =∫_0 ^1 (√(x/(1−x)))ln((x/(1−x)))dx changement (√(x/(1−x)))=t give (x/(1−x)) =t^2 ⇒x =t^2 −t^2 x ⇒(1+t^2 )x =t^2 ⇒ x =(t^2 /(1+t^2 )) =((1+t^2 −1)/(1+t^2 )) =1−(1/(1+t^2 )) ⇒(dx/dt) =((2t)/((1+t^2 )^2 )) ⇒ I =∫_0 ^∞ t ln(((t^2 /(1+t^2 ))/(1−(t^2 /(1+t^2 )))))((2t)/((1+t^2 )^2 ))dt =4 ∫_0 ^∞ (t^2 /((1+t^2 )^2 ))ln(t)dt =4 ∫_0 ^∞ ((1+t^2 −1)/((1+t^2 )^2 ))ln(t)dt =4 ∫_0 ^∞ ((lnt)/(1+t^2 ))dt−4∫_0 ^∞ ((lnt)/((1+t^2 )^2 ))dt =0−4 ∫_0 ^∞ ((lnt)/((1+t^2 )^2 ))dt let A =∫_0 ^∞ ((lnt)/((1+t^2 )^2 ))dt ⇒A =∫_0 ^1 ((lnt)/((1+t^2 )^2 ))dt+∫_1 ^(+∞) ((lnt)/((1+t^2 )^2 ))dt ∫_1 ^(+∞) ((lnt)/((1+t^2 )^2 ))dt =_(t=(1/u)) ∫_0 ^1 ((lnu)/((1+(1/u^2 ))^2 ))×((−du)/u^2 ) =−∫_0 ^1 ((u^2 lnu)/((1+u^2 )^2 ))du ⇒A =∫_0 ^1 (((1−t^2 )lnt)/((1+t^2 )^2 ))dt (1/(1+u)) =Σ_(n=0) ^∞ (−1)^n u^n ⇒ −(1/((1+u)^2 )) =Σ_(n=1) ^∞ n(−1)^n u^(n−1) ⇒ (1/((1+u)^2 )) =Σ_(n=1) ^∞ n(−1)^(n−1) u^(n−1) ⇒ (1/((1+x^2 )^2 )) =Σ_(n=1) ^∞ n(−1)^(n−1) x^(2n−2) ⇒ A =∫_0 ^1 (1−x^2 )lnx(Σ_(n=1) ^∞ n(−1)^(n−1) x^(2n−2) )dx =Σ_(n=1) ^∞ n(−1)^(n−1) ∫_0 ^1 (x^(2n−2) −x^(2n) )lnxdx w_n =∫_0 ^1 (x^(2n−2) −x^(2n) )ln(x)dx =[((1/(2n−1))x^(2n−1) −(1/(2n+1))x^(2n+1) )lnx]_0 ^1 −∫_0 ^1 ((1/(2n−1))x^(2n−2) −(1/(2n+1))x^(2n) )dx =−(1/((2n−1)^2 )) +(1/((2n+1)^2 )) ⇒ A =Σ_(n=1) ^∞ n(−1)^(n−1) ((1/((2n+1)^2 ))−(1/((2n−1)^2 ))) =Σ_(n=1) ^∞ (n/((2n+1)^2 ))(−1)^(n−1) −Σ_(n=1) ^∞ ((n(−1)^(n−1) )/((2n−1)^2 ))(→n=p+1) =Σ_(n=1) ^∞ (n/((2n+1)^2 ))(−1)^(n−1) −Σ_(p=0) ^∞ (((p+1)(−1)^p )/((2p+1)^2 )) =−Σ_(n=1) ^∞ ((n(−1)^n )/((2n+1)^2 )) −Σ_(n=1) ^∞ (((n+1)(−1)^n )/((2n+1)^2 )) −1 =−2 Σ_(n=1) ^∞ ((n(−1)^n )/((2n+1)^2 )) −Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 ))−1 rest to find those sums ...be continued...](https://www.tinkutara.com/question/Q97595.png)