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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

Grade:12

2 Answers

ABHISHEK SINGHAL
18 Points
10 years ago

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
509 Points
10 years ago

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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