Explanation

(d)
$\int_{0}^{1}{\tan^{- 1}{\left( \frac{1}{x^{2} - x + 1} \right)dx}}$
$= \int_{0}^{1}{\tan^{- 1}\left( \frac{x - (x - 1)}{1 + x(x - 1)} \right)}$
$= \int_{0}^{1}{{\tan^{- 1}x}{dx - \int_{0}^{1}{\tan^{- 1}{(x - 1)dx}}}}$
$= \int_{0}^{1}{\tan^{- 1}{x\ dx + \int_{0}^{1}{\tan^{- 1}{(1 - 1 + x)dx}}}}$
$= \int_{0}^{1}{\tan^{- 1}{x\ dx + \int_{0}^{1}{\tan^{- 1}{x\ dx}}}}$
$= 2\int_{0}^{1}{\tan^{- 1}{x\ dx}}$
$= 2\left\lbrack x\tan^{- 1}{x - \int_{}^{}{\frac{x}{1 + x^{2}}dx}} \right\rbrack_{0}^{1}$
$= 2\left\lbrack x\tan^{- 1}{x - \frac{1}{2}\log\left( 1 + x^{2} \right)} \right\rbrack_{0}^{1}$
$= 2\left\lbrack \left\{ 1\tan^{- 1}{1 - \frac{1}{2}\log(2)} \right\} - \left\{ 0 - \frac{1}{2}\log 1 \right\} \right\rbrack$
$= 2\left\lbrack \frac{\pi}{4} - \frac{1}{2}\log{2 - 0} \right\rbrack = \frac{\pi}{2} - \log 2$