IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - Definite Integration good question-2

priyesh

ans is zero

because log((1+x)/(1-x) is an odd function

as log((1-x)/(1+x)) = - log((1+x)/(1-x)

so so I = _{[ -1/2]}

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

=

_{[ -1/2]}

^[1/2]log(1-x/1+x)dx (usinf f(x)dx = f(-1/2 + 1/2 - x) integral prop)

= - _{[ -1/2]}

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

so adding 2i = 0

=> I = 0

Note : for an odd function f(x) definite integral _{[ -a]}^{[a ]} f(x)dx = 0

and 2_{[ 0]}^{[a ]} f(x) dx for an even function