Question Number 149077 by mnjuly1970 last updated on 02/Aug/21

Ω := ∫_0 ^( 1) ((ln ( 1+ (√x) ))/(1+x)) dx =? .....m.n.....

Answered by mindispower last updated on 02/Aug/21

(√x)=t =∫_0 ^1 ((ln(1+t)2tdt)/(1+t^2 ))=A B=∫_0 ^1 ((ln(1−t)2tdt)/(1+t^2 )) A+B=∫_0 ^1 ((ln(1−t^2 )dt^2 )/(1+t^2 ))=∫_0 ^1 ((ln(1−t))/(1+t))dt t→((1−t)/(1+t))=∫_0 ^1 ((ln(((2t)/(1+t))))/(2/(1+t))).(2/((1+t)^2 ))dt =∫_0 ^1 ((ln(2t))/(1+t))−((ln(1+t))/(1+t))dt =(1/2)ln^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−A=∫_0 ^1 ((ln(((1−t)/(1+t))))/(1+t^2 )).2tdt ((1−t)/(1+t))=x =∫_0 ^1 4((ln(x))/((1+x)^2 )).((1−x)/(1+x)).(((1+x)^2 dx)/(2+2x^2 )) =2∫_0 ^1 ((ln(x)(1−x)dx)/((1+x)(1+x^2 ))) ((1−x)/((1+x)(1+x^2 )))=(1/(1+x))−(x/(1+x^2 )) 2∫_0 ^1 ((ln(x))/(1+x))−∫_0 ^1 ((ln(x).dx^2 )/(1+x^2 ))=(3/2)∫_0 ^1 ((ln(x))/(1+x))dx =−(3/2)∫_0 ^1 ((ln(1−(−x)))/(−x))d(−x)=(3/2)Li_2 (−1) A=(1/2)(A+B+A−B) ((ln^2 (2))/4)+((Li_2 (−1))/4)=(1/4)(ln^2 (2)+(π^2 /(12)))

Commented bymnjuly1970 last updated on 02/Aug/21

thanks alot sir power....

Commented byTawa11 last updated on 02/Aug/21

weldone sir.

Commented bymindispower last updated on 03/Aug/21

withe pleasur