Explanation

(d)
Let $I = \int_{0}^{1}{\cot^{- 1}{\left( 1 - x + x^{2} \right)dx}}$
$= \int_{0}^{1}{\tan^{- 1}{\left( \frac{1}{x^{2} - x + 1} \right)dx}}$
$= \int_{0}^{1}{\tan^{- 1}{\left( \frac{x - (x - 1)}{1 + x(x - 1)} \right)dx}}$
$= \int_{0}^{1}{\tan^{- 1}{x\ dx - \int_{0}^{1}{\tan^{- 1}{(x - 1)dx}}\ }}$
$= \int_{0}^{1}{\tan^{- 1}{x\ dx - \int_{0}^{1}{\tan^{- 1}{(1 - x - 1)dx}}}}$
$= 2\int_{0}^{1}{\tan^{- 1}{x\ dx = 2\left\lbrack x\ tan^{- 1}x - \int_{}^{}{\frac{x}{1 + x^{2}}dx} \right\rbrack_{0}^{1}}}$
$= 2\left\lbrack x\tan^{- 1}{x - \frac{1}{2}\log\left( 1 + x^{2} \right)} \right\rbrack_{0}^{1}$
$= 2\left\lbrack \left( 1\tan^{- 1}{1 - \frac{1}{2}\log 2} \right) - \left( 0 - \frac{1}{2}\log 1 \right) \right\rbrack$
$= \frac{\pi}{2} - \log 2$