The integral `int_( -1//2)^(1//2) {[x]+ in ((1+x)/(1-x))}` dx equals

Options

A. `-(1)/(2)`
B. 0
C. 1
D. `2 ln ((1)/(2))`

Answer

A

Explanation

We have,
`underset( -1//2)overset(1//2)int{[x]+ in ((1+x)/(1-x))}dx`
`rArr I=underset( -1//2)overset(1//2)int [x]dx+underset( -1//2)overset(1//2)intln ((1+x)/(1-x))dx`
`rArr I=underset( -1//2)overset(0)int-1dx+underset( -1//2)overset(0)int 0 dx+0 " " [ :. ln ((1+x)/(1-x))"is an odd function"]`
`rArr I=-1(0+(1)/(2))=-(1)/(2)`