integrate xlogx/(1+x2)2 under the limits 0 to infinity

Dear

I = _o∫^∞ xlogx/(1+x²)² dx

first find out

I₁ = ∫x/(1+x²)² dx

     its simple intigration calculate it

I₁ = -1/2(1+x²)

now for I =_o∫^∞ xlogx/(1+x²)² dx

take logx as first function  and x/(1+x²)²  as second function and intigrate by part

I=logx [ -1/2(1+x²)] limit o to ∞      + _o∫^∞ 1/x2(1+x²) dx

I=logx [ -1/2(1+x²)] limit o to ∞      + 1/2 _o∫^∞ x/x²(1+x²) dx

put x²=t in second part

I=logx [ -1/2(1+x²)] limit o to ∞      + 1/4 [ln t/(t+1)] limit o to ∞

now we have to calculate

I =      Lim x→ _∞ [ -logx /2(1+x²)]   - Lim x→ _o  [-logx /2(1+x²)] 

                                                         +Lim t→ _∞ 1/4 [ln t/(t+1)]  - Lim t→ _o 1/4 [ln t/(t+1)]

 third limit is clearly equal to zero

I =      Lim x→ _∞ [ -logx /2(1+x²)]   - Lim x→ _o  [-logx /2(1+x²)]   - Lim t→ _o 1/4 [ln t/(t+1)]

now find out the each limit

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