Given integral, \(I=\int \frac{d x}{x+\sqrt{x-1}}\)
put \(x-1=t^2 \Rightarrow d x=2 t d t\)
then \(I=\int \frac{2 t}{\left(t^2+1\right)+t} d t=\int \frac{(2 t+1)-1}{t^2+t+1} d t\)
\(\begin{aligned}
& =\int \frac{2 t+1}{t^2+t+1} d t-\int \frac{d t}{t^2+t+1} \\
& =\log _e\left|t^2+t+1\right|-\int \frac{d t}{\left(t+\frac{1}{2}\right)^2+\frac{3}{4}} \\
& =\log _e\left|t^2+t+1\right|-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c \\
& =\log _e\left|t^2+t+1\right|-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)+c
\end{aligned}\)
On putting value of \(t=\sqrt{x-1}\), we get
\(I=\log _e|x+\sqrt{x-1}|-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+c\)
Hence, option (4) is correct.