$\int_0^1 \tan ^{-1} x d x=$

To evaluate $\int_0^1 \tan^{-1} x dx$, apply integration by parts with $u = \tan^{-1} x$ and $dv = dx$. Differentiate to find $du = \frac{1}{1+x^2} dx$ and integrate to get $v = x$.

Applying the integration by parts formula yields $\int_0^1 \tan^{-1} x dx = [x \tan^{-1} x]_0^1 - \int_0^1 \frac{x}{1+x^2} dx$. The boundary term evaluates to $[x \tan^{-1} x]_0^1 = (1 \cdot \frac{\pi}{4}) - (0 \cdot 0) = \frac{\pi}{4}$.

The remaining integral simplifies through substitution. Let $t = 1 + x^2$, so $dt = 2x dx$ and $x dx = \frac{1}{2} dt$. When $x = 0$, $t = 1$; at $x = 1$, $t = 2$. The integral becomes $\frac{1}{2} \int_1^2 \frac{1}{t} dt = \frac{1}{2} [\log t]_1^2 = \frac{1}{2} \log 2$.

Combining results gives $\int_0^1 \tan^{-1} x dx = \frac{\pi}{4} - \frac{1}{2} \log 2$, which is equivalent to $\frac{\pi}{4} - \log \sqrt{2}$.

The expression matches option B.