∫ x - 1 ( x + 1 ) x x 2 + x + 1 d x =

Let,

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

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

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

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

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

Let $x+\frac{1}{x}+1=t^2\Rightarrow \left(1-\frac{1}{x^2}\right)dx=2tdt$

$I=\int \left\{\frac{2t}{\left(t^2+1\right)t}\right\}dt$

$=2\int \left\{\frac{1}{\left(t^2+1\right)}\right\}dt$

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

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