Let $I=\int \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) d x$, then $I-2 x \tan ^{-1} x=$

Start with the integral $I = \int \tan^{-1}\left(\frac{2x}{1 - x^2}\right) dx$. Using the identity $2 \tan^{-1} x = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$, valid for $-1 < x < 1$, rewrite the integral as $I = \int 2 \tan^{-1} x dx$.

Apply integration by parts, letting $u = \tan^{-1} x$ and $dv = 2 dx$, so $du = \frac{1}{1 + x^2} dx$ and $v = 2x$. This gives:

$I = 2x \tan^{-1} x - \int \frac{2x}{1 + x^2} dx$

To evaluate the remaining integral, let $t = 1 + x^2$ and $dt = 2x dx$, which yields:

$\int \frac{2x}{1 + x^2} dx = \int \frac{dt}{t} = \log |t| + C = \log(1 + x^2) + C$

Substitute back to find:

$I = 2x \tan^{-1} x - \log(1 + x^2) + C$

The desired expression simplifies as:

$I - 2x \tan^{-1} x = -\log(1 + x^2) + C$

This corresponds to option B.

Answer: B