$$ \int _ {- \infty } ^ { \infty } ( \frac{x}{ a+a^{x}+ a^{-x} } )^{2} dx \wedge \int _ {- \infty } ^ { \infty } ( \frac{x}{a+ e^{ ln a^{x} } + e^{ln a^{-x} } } )^{2} dx \wedge \int _ {- \infty } ^ { \infty } ( \frac{x}{a+ e^{xlna} + e^{-xlna} } )^{2} \Longrightarrow u=xlna \Longrightarrow du=lna.dx \Longrightarrow = \int _ {- \infty } ^ { \infty } \frac{ ( \frac{u}{lna} )^{2} }{ ( a+ e^{u} + e^{-u} )^{2} } . \frac{du}{lna} = \frac{1}{ ln^{3} a} \int _ {- \infty } ^ { \infty } \frac{ u^{2} }{ (a+ e^{u} + e^{-u} )^{2} } du \wedge t= e^{u} \Longrightarrow dt= e^{u} du \Longrightarrow \frac{dt}{t} =du \Longrightarrow = \frac{1}{ ln^{3} a} \int _0^ \infty \frac{t ln^{2}t }{ (a+t+ \frac{1}{t} )^{2} } . \frac{dt}{t}= \frac{a}{ ln^{3}a } \int _0^ \infty \frac{t ln^{2} t}{ ( t^{2} +at+1)^{2} } dt \Longrightarrow a=2:= \frac{1}{ ln^{3} a} \int _0^ \infty \frac{t ln^{2}t }{ (t+1)^{4} } dt$$$$ \Longrightarrow \int _0^ \infty \frac{ x^{s-1} }{ (1+x)^{4} } dx=I(s) \wedge t= \frac{1}{1+x} \Longrightarrow x+1= \frac{1}{t} \Longrightarrow x= \frac{1}{t} -1 \Longrightarrow I(s)= \int _1^0 ( \frac{1}{t} -1)^{s-1} . t^{4} . \frac{-dt}{ t^{2} } = \int _0^1 t^{2} (1-t)^{s-1}. t^{1-s} dt = \int _0^1 t^{3-s} (1-t)^{s-1} dt= \beta (4-s,s) \Longrightarrow \frac{\partial}{\partial s} I(s)= \int _0^ \infty \frac{ x^{s-1} lnx}{ ( 1+x)^{4} } dx= \frac{\partial^2}{\partial s^2} I(s)= \int _0^ \infty \frac{ x^{s-1} ln^{2} x}{ ( 1+x)^{4} } dx \Longrightarrow \frac{\partial^2}{\partial s^2} \beta (4-s,s)= \int _0^ \infty \frac{ x^{s-1} ln^{2}x }{ ( 1+x)^{4} } dx= \frac{\partial^2}{\partial s^2} [ \frac{ \Gamma (4-s) \Gamma (s)}{ \Gamma (4)} ]= \frac{\partial}{\partial s} .[ \frac{1}{ \Gamma (4)} [- \Gamma (s) \Gamma '( 4-s)+ \Gamma (4-s ) \Gamma '(s)]]= \frac{1}{ \Gamma (4)} \frac{\partial}{\partial s} [ \Gamma '(s) \Gamma (4-s)- \Gamma (s) \Gamma '(4-s)]$$
