$\int \frac{x-1}{(x+1)\sqrt{x(x^2+x+1)}}dx=$

$\int \frac{x-1}{(x+1)\sqrt{x(x^2+x+1)}}dx=$

$=\int \frac{x^2-1}{(x+1)\sqrt{x(x^2+x+{\displaystyle \frac{1}{x}})}}dx$

$=\int \frac{x^2-1}{(x^2+2x+1) x\sqrt{x^2+x+\frac{1}{x}}}dx$

$=\int \frac{\left(1-{\displaystyle \frac{1}{x^2}}\right)}{(x^{ }+2+{\displaystyle \frac{1}{x}})\sqrt{x+1+\frac{1}{x}}}dx$

$$\begin{gathered} =\int \frac{\left(2t\right)dt}{\left(t^2+1\right)t} put x+1+\frac{1}{x}=t^2 \\ \Rightarrow \left(1-\frac{1}{x^2}\right)dx=2tdt \end{gathered}$$                 

$=2\int \frac{dt}{t^2+1}$

$=2 {\tan }^{-1}t+C$

$=2 {\tan }^{-1}\left(\sqrt{x+1+\frac{1}{x}}\right)+C$