∫(от 1/6 до 2) (x / (3x2 – x + 1)) dx
[m]3x^2-x+1=3\cdot (x^2-\frac{1}{3}x+\frac{1}{3})=3\cdot (x-\frac{1}{6})^2+\frac{11}{36})[/m]
Замена переменной:
[m]x-\frac{1}{6}=t[/m]
[m]x= t + \frac{1}{6} [/m]
[m] dx=dt[/m]
Пределы:
[m]x=\frac{1}{6}[/m] ⇒ [m]t=0[/m]
[m]x=2[/m] ⇒ [m]t=\frac{11}{6}[/m]
[m]\int_{\frac{1}{6}}^{2}\frac{xdx}{3x^2-x+1}=\frac{1}{3}\int_{0}^{\frac{11}{6}}\frac{(t+\frac{1}{6})dt}{t^2+\frac{11}{36}}=\frac{1}{6}\int_{0}^{\frac{11}{6}}\frac{2tdt}{t^2+\frac{11}{36}}+\frac{1}{18}\int_{0}^{\frac{11}{6}}\frac{dt}{t^2+\frac{11}{36}}=[/m]
[m]=\frac{1}{6}ln|t^2+\frac{11}{36}||_{0}^{\frac{11}{6}}+\frac{1}{18}\cdot \frac{1}{\sqrt{\frac{11}{36}}}\cdot arctg \frac{t}{\sqrt{\frac{11}{36}}}|_{0}^{\frac{11}{6}}= [/m]
считайте...