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vishal .... Grade: 12

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

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points


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

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Askiitians Experts

7 years ago
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