Let $I=\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} d x$
$\begin{aligned} & =\int \frac{(x-1)(x+1)}{(x+1)(x+1) \sqrt{x^3+x^2+x}} d x \\ & =\int \frac{x^2-1}{(x+1)^2 \sqrt{x^3+x^2+x}} d x \\ & =\int \frac{x^2-1}{\left(x^2+2 x+1\right) \sqrt{x^3+x^2+x}} d x \\ & =\int \frac{x^2\left(1-1 / x^2\right) d x}{x\left(x+2+\frac{1}{x}\right) \cdot x \sqrt{x+1+\frac{1}{x}}} \\ & =\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}+2\right) \sqrt{x+\frac{1}{x}+1}}\end{aligned}$
Put, $x+\frac{1}{x}+1=t^2 \Rightarrow\left(1-\frac{1}{x^2}\right) d x=2 t d t$
$\begin{aligned} \therefore \quad I & =\int \frac{2 t d t}{\left(t^2+1\right) t}=2 \int \frac{d t}{t^2+1}=2 \tan ^{-1} t \\ & =2 \tan ^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)+C\end{aligned}$
$\therefore \quad f(x)=2 \tan ^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)$
$\therefore \quad f(1)=2 \tan ^{-1} \sqrt{3}=2 \times \frac{\pi}{3}=\frac{2 \pi}{3}$