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`        integrate   xlogx/(1+x2)2 under the limits 0 to infinity`
7 years ago

147 Points
```										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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```
7 years ago
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