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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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