I=int_(0)^(1)sin^(-1)[(2x)/(1+x^(2))]dx

Find the values of int_(0)^(1)sin^(-1).(2x)/(1+x^(2))dx(-1lexle1)

int_(-1)^(1)sin^(-1).(2x)/(1+x^(2))dx=0

int_(0)^(1/2)(sin^(-1)x)/((1-x^(2))^((1)/(2)))dx=

int_(0)^(1)(d)/(dx)["sin"^(-1)(2x)/(1+x^(2))]dx is equal to -

I=int_(0)^(1)((1+x)/(2-x))dx

If I_(1) int sin^(-1) ((2x)/(1 +x^(2)) ) dx , I_(2) = int cos^(-1) ((1-x^(2))/(1 +x^(2)) ) dx , I_(3) = int tan^(-1) ((2x)/(1 - x^(2)) ) dx , then I_(1) + I_(2) - I_(3) =

If I_(1)=int_(0)^((pi)/(2))(x)/(sin x)dx and I_(2)=int_(0)^(1)(tan^(-1)x)/(x)dx, then (I_(1))/(I_(2))=(A)1(B)(1)/(2) (C) 2 (D) (pi)/(2)

If I_(1)= int Sin^(-1)((2x)/(1+x^(2)))dx,I_(2)= int Cos^(-1)((1-x^(2))/(1+x^(2)))dx, I_(3)= intTan^(-1)((2x)/(1-x^(2)))dx then the value of I_(1)+I_(2)-I_(3)=