$\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x=$

$\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^2}\right)+C$

We have, $2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

∴  $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}=\frac{1}{2} \cos ^{-1}\left(\frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}\right)=\frac{1}{2} \cos ^{-1} x$

Thus, we have

$I=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$

$\Rightarrow I=\frac{1}{2} \int \cos ^{-1} x d x$

$\Rightarrow I=\frac{1}{2}\left\{x \cos ^{-1} x-\int \frac{-1}{\sqrt{1-x^2}} \times x d x\right\}$

$\Rightarrow I=\frac{1}{2}\left\{x \cos ^{-1} x-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^2}} d x\right\}$

$\Rightarrow I=\frac{1}{2}\left\{x \cos ^{-1} x-\sqrt{1-x^2}\right\}+C$