$\int_0^1 tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$ is equal to

Looking at the expression inside: $\frac{2x-1}{1+x-x^2}$

The formula for $\tan(A - B)$ is:

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

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Let $\tan A = x$ and $\tan B = 1-x$

The numerator becomes:

$\tan A - \tan B = x - (1-x)$

$= x - 1 + x$

$= 2x - 1$

The denominator becomes:

$1 + \tan A \tan B = 1 + x(1-x)$

$= 1 + x - x^2$

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Since $\frac{2x-1}{1+x-x^2} = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:

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

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The integral becomes:

$\int_0^1 [\tan^{-1}(x) - \tan^{-1}(1-x)]dx$

$= \int_0^1 \tan^{-1}(x)dx - \int_0^1 \tan^{-1}(1-x)dx$

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For $\int_0^1 \tan^{-1}(1-x)dx$, let $u = 1-x$

Then $du = -dx$

When $x = 0$: $u = 1$

When $x = 1$: $u = 0$

$\int_0^1 \tan^{-1}(1-x)dx = -\int_1^0 \tan^{-1}(u)du$

$= \int_0^1 \tan^{-1}(u)du$

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$\int_0^1 \tan^{-1}(x)dx - \int_0^1 \tan^{-1}(1-x)dx = \int_0^1 \tan^{-1}(x)dx - \int_0^1 \tan^{-1}(x)dx$

$= 0$

Therefore, the value of the integral is $0$.