Correct Answer - `theta cos 2 theta-(sin2 theta)/(2)+C`
` I=int tan^(-1)sqrt((1-x)/(1+x))dx`
` "Let "x=cos2 theta " or "dx=-2sin2 theta d theta `
`:. I=int tan^(-1)sqrt((1-cos2 theta)/(1+cos2 theta))(-2sin2 theta)d theta `
`=-2int tan^(-1)(tan theta)sin2 theta d theta `
`=-2int theta sin 2 theta d theta `
`=-2[-(theta cos2 theta)/(2)+int(cos2 theta)/(2) d theta]`
`=theta cos 2 theta-(sin2 theta)/(2)+C`
