[log(1+x)]/(1+x^2) lower lim 0 upper lim 1

Question

ABHISHEK SINGHAL · Answer

PUT X=TANT ASSUME FUNCTUION TO BE I AND THE LIMITS OF INTEGRATION WOULD BE 0 TO PIE/4 USE F(A+B-X)=F(X) AND 1ST AND SECOND U WILL SURE GET ANS I,E (PIE/8)LN2

vikas askiitian expert · Answer

I = [log(1+x)] dx / (1+x 2 ) lim 0 to 1 put log(1+x) = t dx = (1+x)dt = e t dt I = [t] e t d t /(1+(e t - 1) 2 ) lim 0 to log2 I = [t] e t dt /(1+(e t -1) 2 ) lim 0 to 0.3 [t] , greatest integer t value is equal to 0 for this interval ... I = 0

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[log(1+x)]/(1+x^2) lower lim 0 upper lim 1

[log(1+x)]/(1+x^2)

lower lim 0

upper lim 1

2 Answers

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