0-1-x-tanh-1-x-1-x-2-dx-1-24-pi-2-6- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 176542 by mnjuly1970 last updated on 20/Sep/22 Ω=∫01x.tanh−1(x)(1+x)2dx=124(π2−6) Answered by Peace last updated on 20/Sep/22 ∫01x(1+x)2tg−(x)dxtanh−(x)=12(ln(1+x)−ln(1−x))x(1+x)2=11+x−1(1+x)22Ω=∫01ln(1+x)1+x−ln(1+x)(1+x)2dx+∫01ln(1−x)(1+x)2dx−∫01ln(1−x)1+xdx=12[ln2(1+x)]01+[ln(1+x)1+x]01−∫011(1+x)2+A+B=ln2(2)2+ln(2)2+12+A+BA=limx→1∫0xln(1−t)(1+t)2dt=limx→1{−ln(1−x)1+x−∫0x11−t2.dt}=limx→1−ln(1−x)1+x−12∫0x11−t+11+tdt=limx→1−ln(1−x)1+x+ln(1−x)2−ln(1+x)2=−ln(2)2−B=∫01ln(1−x)1+xdxx=1−t1+t,=−1+21+t⇒dt=−2dt(1+t)22∫01ln(2t1+t)2(1+t).1(1+t)2dt=∫01ln(2t)−ln(1+t)1+tdt=∫01ln(2)1+t+ln(t)1+t−ln(1+t)1+tdt=ln2(2)−ln2(2)2+∫01ln(t)1+tdt=ln2(2)2+∫01ln(1−(−t)−td(−t)=ln2(2)2−Li2(−1)B=−ln2(2)2+Li2(−1)2Ω=ln2(2)2+ln(2)2+12+A+B=ln2(2)2+ln(2)2+12−ln(2)2−ln2(2)2+Li2(−1)=12+Li2(−1)=12+π212=112(π2−6)Ω=124(π2−6) Commented by Tawa11 last updated on 22/Sep/22 Greatsir Commented by Peace last updated on 23/Sep/22 thankYouniceDay Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Differentiate-with-respect-to-x-arctan-a-2-x-2-a-2-x-2-Next Next post: Question-176540 Leave a Reply Cancel replyYour email address will not be published.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.
![∫_0 ^1 (x/((1+x)^2 ))tg^− (x)dx tanh^− (x)=(1/2)(ln(1+x)−ln(1−x)) (x/((1+x)^2 ))=(1/(1+x))−(1/((1+x)^2 )) 2Ω=∫_0 ^1 ((ln(1+x))/(1+x))−((ln(1+x))/((1+x)^2 ))dx+∫_0 ^1 ((ln(1−x))/((1+x)^2 ))dx−∫_0 ^1 ((ln(1−x))/(1+x))dx =(1/2)[ln^2 (1+x)]_0 ^1 +[((ln(1+x))/(1+x))]_0 ^1 −∫_0 ^1 (1/((1+x)^2 ))+A+B =((ln^2 (2))/2)+((ln(2))/2)+(1/2)+A+B A=lim_(x→1) ∫_0 ^x ((ln(1−t))/((1+t)^2 ))dt=lim_(x→1) {−((ln(1−x))/(1+x))−∫_0 ^x (1/(1−t^2 )).dt} =lim_(x→1) −((ln(1−x))/(1+x))−(1/2)∫_0 ^x (1/(1−t))+(1/(1+t))dt =lim_(x→1) −((ln(1−x))/(1+x))+((ln(1−x))/2)−((ln(1+x))/2) =−((ln(2))/2) −B=∫_0 ^1 ((ln(1−x))/(1+x))dx x=((1−t)/(1+t)),=−1+(2/(1+t))⇒dt=−((2dt)/((1+t)^2 )) 2∫_0 ^1 ((ln(((2t)/(1+t))))/(2/((1+t)))).(1/((1+t)^2 ))dt=∫_0 ^1 ((ln(2t)−ln(1+t))/(1+t))dt =∫_0 ^1 ((ln(2))/(1+t))+((ln(t))/(1+t))−((ln(1+t))/(1+t))dt =ln^2 (2)−((ln^2 (2))/2)+∫_0 ^1 ((ln(t))/(1+t))dt=((ln^2 (2))/2)+∫_0 ^1 ((ln(1−(−t))/(−t))d(−t) =((ln^2 (2))/2)−Li_2 (−1) B=−((ln^2 (2))/2)+Li_2 (−1) 2Ω=((ln^2 (2))/2)+((ln(2))/2)+(1/2)+A+B =((ln^2 (2))/2)+((ln(2))/2)+(1/2)−((ln(2))/2)−((ln^2 (2))/2)+Li_2 (−1) =(1/2)+Li_2 (−1)=(1/2)+(π^2 /(12))=(1/(12))(π^2 −6) Ω=(1/(24))(π^2 −6)](https://www.tinkutara.com/question/Q176545.png)
