`int_0^1cot ^(-1) (1-x +x^2) dx` = (1)`pi/2 - log 2` `(2) pi/2+ log 2` `(3)pi- log 2` `(4) pi + log 2`

int_(0)^(1)cot^(-1)(1-x+x^(2))dx=(1)(pi)/(2)-log2(2)(pi)/(2)+log2(3)pi-log2(4)pi+log2

The value of int_(0)^(1)(8log(1+x))/(1+x^(2))dx is: (1)pi log2 (2) (pi)/(8)log2(3)(pi)/(2)log2(4)log2

int_(0)^(2pi) ln (1+Cos x) dx=

int _(0) ^(pi//2) (cos x )/( 1 + sin x ) dx equals to a) log 2 b) 2 log 2 c) (log2) ^(2) d) 1/2 log 2

int_(0)^(1) ln Sin((pi x)/(2))dx=

[int_(0)^(0)cot^(-1)(1-x+x^(2))dx=],[[" (A) "(pi)/(2)-ln2," (B) "(pi)/(2)+ln2],[" (C) "(pi)/(2)-ln sqrt(2)," (D) "(pi)/(2)+ln sqrt(2)]]

The value of int_0^1(8log(1+x))/(1+x^2)dx is a. pilog2 b. \ pi/8log2 c. \ pi/2log2 d. log2

int_(0)^(1)(log|1+x|)/(1+x^(2))dx=(pi)/(8)log2

If int_(0)^(pi//2) log(cosx) dx=pi/2 log (1/2), then int_(0) ^(pi//2) log (sec x ) dx =