If y = ^ - 1 ( 1 x ^ 2 + x + 1 ) + ^ - 1 ( 1 x ^ 2 + 3x + 3 ) + ta...

$y = \tan^{- 1}\left(\dfrac{1}{{x}^{2} + x + 1}\right) + \tan^{- 1}\left(\dfrac{1}{{x}^{2} + 3x + 3}\right)$
$+ \tan^{- 1} ( \dfrac{1}{{x}^{2} + 5x + 7} ) + \tan^{- 1}\left(\dfrac{1}{{x}^{2} + 7x + 13}\right)$
$= \tan^{- 1}\left(\dfrac{1}{x\left(x + 1\right) + 1}\right) + \tan^{- 1}\left(\dfrac{1}{\left(x + 2\right)\left(x + 1\right) + 1}\right) + \tan^{- 1}\left(\dfrac{1}{\left(x + 3\right)\left(x + 2\right) + 1}\right)c + \tan^{- 1}\left(\dfrac{1}{\left(x + 4\right)\left(x + 3\right) + 1}\right)$
$= \tan^{- 1}\left(\dfrac{\left(x + 1\right) - x}{1 + \left(x + 1\right)x}\right) + \tan^{- 1}\left(\dfrac{\left(x + 2\right) - \left(x + 1\right)}{1 + \left(x + 2\right)\left(x + 1\right)}\right)$+ \tan^{- 1}\left(\dfrac{\left(x + 3\right) - \left(x + 2\right)}{1 + \left(x + 3\right)\left(x + 2\right)}\right) + \tan^{- 1}\left(\dfrac{\left(x + 4\right) - \left(x + 3\right)}{1 + \left(x + 4\right)\left(x + 3\right)}\right)$<br />$= \tan^{- 1}\left(x + 1\right) - \tan^{- 1}x + \tan^{- 1}\left(x + 2\right)$- \tan^{- 1}\left(x + 1\right) + \tan^{- 1}\left(x + 3\right) - \tan^{- 1}\left(x + 2\right) + \tan^{- 1}\left(x + 4\right) - \tan^{- 1}\left(x + 3\right)$
$⇒ y = \tan^{- 1}\left(x + 4\right) - \tan^{- 1}x$
$⇒ \dfrac{dy}{dx} = \dfrac{1}{1 + {( x + 4 )}^{2}} - \dfrac{1}{1 + {x}^{2}}$
$⇒ {\left(\dfrac{dy}{dx}\right)}_{x = 0} = \dfrac{1}{1 + 16} - 1$
$= - \dfrac{16}{17}$
$= \dfrac{- 16}{1 + 16}$
$⇒ k = 16$