[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

I = [log(1+x)] dx / (1+x²)          lim 0 to 1

put log(1+x) = t

dx = (1+x)dt = e^tdt

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