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Question Number 96365 by Ar Brandon last updated on 31/May/20

Answered by  M±th+et+s last updated on 01/Jun/20

x=(1/t) dx=−(dt/t^2 ) I=∫_∞ ^0 (((1/t^2 ).(−ln(t)).(−(dt/t^2 )))/((((1+t^2 ))/t^2 ).(((1+t)^2 )/t^2 ))) I=−∫_0 ^∞ ((ln(x))/((1+x^2 )(1+x)^2 ))dx....2 2I=∫_0 ^∞ ((ln(x)[x^2 −1])/((1+x^2 )(1+x)^2 ))dx ∫_0 ^∞ =∫_0 ^1 +∫_1 ^∞ ((ln(x)[x^2 −1])/((1+x)^2 (1+x^2 )))dx x=(1/u) dx=((−du)/u^2 ) ∫_1 ^0 ((−ln(u).(([1−u^2 ])/u^2 ) .((−du)/u^2 ))/((((1+u)^2 )/u^2 ).(((1+u^2 ))/u^2 ))) 2I=∫_0 ^1 ((ln(u)(u^2 −1))/((1+u)^2 (u^2 +1)))du I=∫_0 ^1 (((u−1)ln(u))/((u+1)(u^2 +1)))du I=∫_0 ^1 ln(u)[(u/(u^2 +1))−(1/(u+1))]du I=∫_0 ^1 ((uln(u))/(u^2 +1))du−∫_0 ^1 ((ln(u))/(u+1))du i will continue later ......

x=1tdx=−dtt2I=∫∞01t2.(−ln(t)).(−dtt2)(1+t2)t2.(1+t)2t2I=−∫0∞ln(x)(1+x2)(1+x)2dx....22I=∫0∞ln(x)[x2−1](1+x2)(1+x)2dx∫0∞=∫01+∫1∞ln(x)[x2−1](1+x)2(1+x2)dxx=1udx=−duu2∫10−ln(u).[1−u2]u2.−duu2(1+u)2u2.(1+u2)u22I=∫01ln(u)(u2−1)(1+u)2(u2+1)duI=∫01(u−1)ln(u)(u+1)(u2+1)duI=∫01ln(u)[uu2+1−1u+1]duI=∫01uln(u)u2+1du−∫01ln(u)u+1duiwillcontinuelater......

Answered by  M±th+et+s last updated on 01/Jun/20

I_1 =∫_0 ^1 ((uln(u))/(u^2 +1))du=∫_0 ^1 ln(u) d((1/2)ln∣u^2 +1∣) =[(1/2)ln(u)ln(u^2 +1)]_0 ^1 −(1/2)∫_0 ^1 ((ln(1+u^2 ))/u)du =−(1/2)∫_0 ^1 Σ_(n=1) ^∞ (((−1)^(n+1) )/n).u^(2n−1) du =−(1/2)Σ_(n=1) ^∞ (((−1)^(n+1) )/n)∫_0 ^1 u^(2n−1) du =−(1/4)Σ_(n=1) ^∞ (((−1)^(n+1) )/n^2 )=((−π^2 )/(48)) I_2 =∫_0 ^1 ((ln(u))/(1+u))du=−(1/2)((π^2 /6))=((−π^2 )/(12)) I=I_1 −I_2 I=−(π^2 /(48))+(π^2 /(12))=(π^2 /(16)) hence proved

I1=∫01uln(u)u2+1du=∫01ln(u)d(12ln∣u2+1∣)=[12ln(u)ln(u2+1)]01−12∫01ln(1+u2)udu=−12∫01∑∞n=1(−1)n+1n.u2n−1du=−12∑∞n=1(−1)n+1n∫01u2n−1du=−14∑∞n=1(−1)n+1n2=−π248I2=∫01ln(u)1+udu=−12(π26)=−π212I=I1−I2I=−π248+π212=π216henceproved

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