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[log(1+x)]/(1+x^2) lower lim 0 upper lim 1
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
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
I = [log(1+x)] dx / (1+x2) lim 0 to 1 put log(1+x) = t dx = (1+x)dt = etdt I = [t] et d t /(1+(et - 1)2) lim 0 to log2 I = [t] et dt /(1+(et-1)2) lim 0 to 0.3 [t] , greatest integer t value is equal to 0 for this interval ... I = 0
I = [log(1+x)] dx / (1+x2) lim 0 to 1
put log(1+x) = t
dx = (1+x)dt = etdt
I = [t] et d t /(1+(et - 1)2) lim 0 to log2
I = [t] et dt /(1+(et-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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