Explanation

(c)
Let $I = \int_{}^{}\frac{dx}{x + \sqrt{x - 1}}$
Put $x = t^{2} + 1\ \ \Rightarrow \ \ dx = 2t\ dt$
$\therefore\ I = \int_{}^{}\frac{2t}{t^{2} + t + 1}dt$
$= \int_{}^{}\frac{2t + 1}{t^{2} + t + 1}dt - \int_{}^{}\frac{1}{t^{2} + t + 1}dt$
$= \log{(t^{2} + t + 1)} - \int_{}^{}\frac{1}{\left( t + \frac{1}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2}}dt$
$= \log{\left( t^{2} + t + 1 \right) - \frac{2}{\sqrt{3}}}\tan^{- 1}\left( \frac{2t + 1}{\sqrt{3}} \right) + c$
$= \log{\left( x + \sqrt{x - 1} \right) - \frac{2}{\sqrt{3}}}\tan^{- 1}\left( \frac{2\sqrt{x - 1} + 1}{\sqrt{3}} \right) + c$