The value of \(\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x\) is

\(\text {Let } I=\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x \)
\( =\int_0^1 \tan ^{-1}\left[\frac{x+(x-1)}{1+x(1-x)}\right] d x=\int_0^1\) \(\tan ^{-1}\left[\frac{x+(x-1)}{1-(x-1)(x)}\right] d x \)
\( =\int_0^1[\tan ^{-1} x+\tan ^{-1}(x-1) d x=\int_0^1 \tan ^{-1} x d x\) \(+\int_0^1 \tan ^{-1}(x-1) d x\)
\( \left.=[\left[x \tan ^{-1} x\right]_0^1-\frac{1}{2} \int_0^1 \frac{2 x}{1+x^2} d x\right]+\left[x \tan ^{-1}(x-1)\right]_0^1\) \(-\frac{1}{2} \int_0^1 \frac{2 x-2+2}{1+(x-1)^2} d x] \)
\( =\left(\frac{\pi}{4}\right)-\frac{1}{2}\left[\log \left|1+x^2\right|\right]_0^1+0-\frac{1}{2} \int_0^1 \frac{2 x-2}{1+(x-1)^2}\) \( d x-\int_0^1 \frac{d x}{1+(x-1)^2} \)
\( =\left(\frac{\pi}{4}\right)-\frac{1}{2} \log 2-\frac{1}{2}\left[\log \left|1+(x-1)^2\right|\right]_0^1-\) \(\left[\tan ^{-1}(x-1)\right]_0^1 \)
\( =\frac{\pi}{4}-\frac{1}{2} \log 2-\frac{1}{2}(0-\log 2)-\left[0-\tan ^{-1}(-1)\right] \)
\( =\frac{\pi}{4}-\frac{1}{2} \log 2+\frac{1}{2} \log 2-\frac{\pi}{4}=0\)