The value of ∫ _ 0 ^ 1 ^ -1 (1 - x + x^ 2 )dx is

$\cot^{-1} \left(1 - x + x^{2}\right) = \tan^{-1} \left(\dfrac{1}{1 - x + x^{2}}\right)</p><p>$⇒ \cot^{-1} \left(1 - x + x^{2}\right) =\tan^{-1} \left(\dfrac{1}{1 - x\left(1 - x\right)}\right)

⇒ \cot^{-1} \left(1 - x + x^{2}\right) =\tan^{-1} \left(\dfrac{x + \left(1 - x\right)}{1 - x\left(1 - x\right)}\right)</p><p>⇒ \cot^{-1} \left(1 - x + x^{2}\right) = \tan^{-1}x + \tan^{-1}\left(1 - x\right)$</p><p>$∴ {∫}_{0}^{1}\cot^{-1}\left(1 - x + x^{2}\right)dx = {∫}_{0}^{1}\tan^{-1}xdx + {∫}_{0}^{1}\tan^{-1}$\left(1 - x\right)dx$

$⇒ {∫}_{0}^{1}\cot^{-1}\left(1 - x + x^{2}\right)dx = {∫}_{0}^{1}\tan^{-1}xdx + {∫}_{0}^{1}\tan^{-1}xdx$ & \left[∵ {∫}_{0}^{a}f \left(x\right)$dx = {∫}$_{0}^{a}f\left(a - x\right)dx\right]$</p><p>$⇒ {∫}_{0}^{1}\cot^{- 1}\left(1 - x + x^{2}\right)$dx = 2{∫}_{0}^{1}\tan^{-1}xdx$

On evaluating by integration by parts, we have

$⇒ {∫}_{0}^{1}\cot^{- 1}$\left(1 - x + x^{2}\right)dx = 2\left\{{\tan^{-1}x \cdotx}_{0}^{1} - {∫}_{0}^{1}\dfrac{ x}{1 + x^{2}}dx\right\}$</p><p>$⇒ {∫}_{0}^{1}\cot^{- 1}\left(1 - x + x^{2}\right) dx = 2 \left\${\dfrac{\pi}{4} -{\dfrac{1}{2}\ln\left(1 + x^{2}\right)}_{0}^{1}\right\}$</p><p>$⇒ {∫}_{0}^{1}\cot^{-1}\left(1 - x + x^{2}\right)dx = 2\left[\dfrac{\pi}{4} - \dfrac{1}{2}\log_{e}2\right] = \dfrac{\pi}{2} -\log_{e}2

Hence, ${∫}_{0}^{1}\cot^{-1}\left(1 - x + x^{2}\right)dx = \dfrac{\pi}{2} -\log_{e}2$$</p><p>So, option$( b )$is correct.</p>$