$\frac{x}{(x+1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}=\frac{A(x^2+x+1)+(Bx+C)(x-1)}{(x-1)(x^2+x+1)}=>\\=>x=A(x^2+x+1)+(Bx+C)(x-1);\\x=Ax^2+Ax+A+Bx^2-Bx+Cx-C=(A+B)x^2+(A-B+C)x+A-C=>\\=>\left \{ {{A+B=0} \atop {A-B+C=1}}\atop{A-C=0} \right.\left \{ {{A=-B} \atop {A-B+C=1}}\atop{A=C} \right.;\left \{ {{-B-B+C=1} \atop {-B-C=0}} \right.;-3B=1;B=-\frac{1}{3}=>A=C=\frac{1}{3}$

Запишем данный интеграл, как сумму интегралов

$\int{\frac{x}{(x+1)(x^2+x+1}dx}=\frac{1}{3}\int{\frac{1}{x-1}dx}+\int{\frac{-\frac{1}{3}x+\frac{1}{3}}{x^2+x+1}dx}$

Посчитаем эти интегралы

$1)\frac{1}{3}\int{\frac{1}{x-1}dx}=\frac{1}{3}\ln(|x-1|)+C\\2)\int{\frac{-\frac{1}{3}x+\frac{1}{3}}{x^2+x+1}dx}=\begin{vmatrix}Px+Q=(x^2+x+1)'=\\=-\frac{1}{6}(2x-2)\end{vmatrix}=-\frac{1}{6}\int{\frac{2x-2}{x^2+x+1}dx}=-\frac{1}{6}\int{\frac{d(x^2+x+1)}{x^2+x+1}}+\frac{1}{2}\int{\frac{1}{x^2+x+1}dx(1)}=-\frac{1}{6}\ln(x^2+x+1)+\frac{2}{\sqrt{3}}\arctan(\frac{2(x+\frac{1}{2})}{\sqrt{3}})+C$

$(1)\\\int{\frac{1}{x^2+x+1}dx=\int\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}=\frac{2}{\sqrt{3}}\arctan(\frac{2(x+\frac{1}{2})}{\sqrt{3}})+C$

Объединяем решения

$\frac{1}{3}\ln(|x-1|)-\frac{1}{6}\ln(|x^2+x+1|)+\frac{1}{\sqrt{3}}\arctan(\frac{2x+1}{\sqrt{2}})+C=\\=\frac{1}{3}(]ln(]x-1|)-\frac{1}{2}\ln(x^2+x+1))+\frac{1}{\sqrt{3}}\arctan(\frac{2x+1}{\sqrt{2}})+C=\\=\frac{1}{3}\ln(\frac{|x-1|}{\sqrt{x^2+x+1}})+\frac{1}{\sqrt{3}}\arctan(\frac{2x+1}{\sqrt{2}})+C$

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4 года назад