∫_0^1 tan^-1≤ft((2x-1)/(1+x-x^2)) dx is equal to

The correct answer is Option (2) → 0

Let I be the integral. By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\) with a = 1, x is replaced by 1 − x:

    \(2(1 - x) - 1 = 1 - 2x\)
    \(1 + (1 - x) - (1 - x)^2 = 1 + x - x^2\)

So the numerator changes sign and the denominator is unchanged:

    \( \displaystyle I = \int_{0}^{1} \tan^{-1}\left(\frac{1 - 2x}{1 + x - x^2}\right) dx \)

Since \(\tan^{-1}(-t) = -\tan^{-1}t\) (odd function):

    \( \displaystyle I = -\int_{0}^{1} \tan^{-1}\left(\frac{2x - 1}{1 + x - x^2}\right) dx = -I \)

    \(2I = 0,\quad I = 0\)