`int_(-1)^(1)log[(a-x)/(a+x)]dx`

int_(-1)^(1)log((4-x)/(4+x))dx=

I=int_(-1)^(1)log((2-x)/(2+x))dx

int_(0)^(1)(log(1-x))/(x)dx

int_(-1)^(1)log((1+x)/(1-x))dx=

int_(-1)^1 log((3-x)/(3+x)) dx

The value of int_(-1)^(1) log ((x-1)/(x+1))dx is

Evaluate the following integral: int_(-1)^1log((2-x)/(2+x))dx\

Prove that int_-a^a f(x) dx=0 , where 'f' is an odd function. And, evaluate, int_-1^1 log[(2-x)/(2+x)] dx

int_(0)^(1)(log(1+x))/(1+x)dx

int_0^pi dx/(1+10^(cosx))+int_(-1)^1 log((2-x)/(2+x))dx= (A) pi/2 (B) -pi (C) 0 (D) none of these