\(\begin{aligned}
& \int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} d x \\
& \quad=\int \frac{x^2-1}{(x+1)^2 x \sqrt{1+\frac{1}{x}+x}} d x \\
& \quad=\int \frac{x^2-1}{\left(1+1+\frac{1}{x}+x\right) x^2 \sqrt{1+\frac{1}{x}+x}} d x \\
& \quad=\int \frac{1}{\left[1+\left(1+\frac{1}{x}+x\right)\right] \sqrt{1+\frac{1}{x}+x}} d x
\end{aligned}\)
Let \(1+\frac{1}{x}+x=t^2\)
\(\begin{aligned}
& \Rightarrow\left(1-\frac{1}{x^2}\right) d x=2 t d t=\int \frac{2 t d t}{\left(1+t^2\right) t}=2 \int \frac{d t}{1+t^2} \\
& =2 \tan ^{-1}(t)+c=2 \tan ^{-1}\left(\sqrt{\frac{1+x+x^2}{x}}\right)+C
\end{aligned}\)
Hence, option (1) is correct.