[m]3x^2+x+1=3(x+\frac{1}{3}x+\frac{1}{3})=3((x+\frac{1}{6})^2-\frac{1}{36}+\frac{1}{3})=3((x+\frac{1}{6})^2+\frac{11}{36})[/m]
Замена переменной:
[m]x+\frac{1}{6}=t[/m]
[m]dx=dt[/m]
получаем
[m] ∫ \frac{x+6}{3x^2+x+1}dx= ∫\frac{x+6}{3((x+\frac{1}{6})^2+\frac{11}{36})} dx= ∫ \frac{t-\frac{1}{6}+6}{3(t^2+\frac{11}{36})} dt=[/m]
[m]= \frac{1}{3} ∫ \frac{t}{t^2+\frac{11}{36}}dt-\frac{35}{18} ∫ \frac{1}{t^2+\frac{11}{36}}dt=\frac{1}{6}ln|t^2+\frac{11}{36}|-\frac{35}{18}\cdot \frac{1}{ \sqrt{\frac{11}{36}}}arctg\frac{t}{\sqrt{\frac{11}{36}}} +C=\frac{1}{6}ln|x^2+\frac{1}{3}x+\frac{1}{3}|-\frac{35}{3 \sqrt{11}}arctg\frac{6x+1}{\sqrt{11}} +C[/m]