\int \ arcctg{\left( x^2+x+1 \right )} dx=\\
\left [ \begin {array}{cc}u=\operatorname{arccot}\left( x^2+x+1 \right )&dv=dx\\du=-\frac{2x+1}{\left( x^2+x+1 \right )^2+1}dx&v=x \end{array} \right ]=\\
x\ arccot\left ( x^2+x+1 \right )+\int \frac{2x^2+x}{\left ( x^2+x+1 \right )^2+1}dx=\\
x\ arccot\left ( x^2+x+1 \right )+\int \frac{2x^2+x}{\left(x^2+1\right) \left(x^2+2 x+2\right)}dx=\\
x\ arccot\left ( x^2+x+1 \right )+\int \frac{x}{x^2+1}-\frac{x}{x^2+2 x+2}dx=\\
x\ arccot\left ( x^2+x+1 \right )+\frac{1}{2}\ln{\left (x^2+1\right )}-\int \frac{x}{x^2+2 x+2}dx=\\
x\ arccot\left ( x^2+x+1 \right )+\frac{1}{2}\ln{\left (x^2+1\right )}-\int \frac{x+1}{x^2+2 x+2}dx+\int \frac{1}{(x^2+2x+1)+1}=\\
x\ arccot\left ( x^2+x+1 \right )+\frac{1}{2}\ln{\left (x^2+1\right )}-\frac{1}{2}\ln{\left (x^2+2x+2 \right )}+\int \frac{1}{(x+1)^2+1}=\\
x\ arccot\left ( x^2+x+1 \right )+\frac{1}{2}\ln{\left (x^2+1\right )}-\frac{1}{2}\ln{\left (x^2+2x+2 \right )}+\ arctg\left (x+1 \right )+C