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

148 Points
14 years ago

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)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

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

All the best.

Regards,