Интеграл..............................

Решите задачу:

$\int {\frac{2x^2 -1}{x^3+6x^2 +9x}} \, dx = \int {\frac{2x^2 -1}{x \cdot (x^2+6x +9)}} \, dx = \int {\frac{2x^2 -1}{x \cdot (x+3)^2}} \, dx = (*) \\ \\ \frac{A}{x} + \frac{B}{x+3}+ \frac{C}{(x+3)^2}=\frac{2x^2-1}{x \cdot (x+3)^2} \\ \\ A(x^2 +6x+9) + B(x^2+3x) +Cx=2x^2-1$

$\left\{\!\begin{aligned} & A +B=2 \\ & 6A+3B+C=0 \\ & 9A =-1 \end{aligned}\right. \ \ \left\{\!\begin{aligned} & B = 2+\frac{1}{9}=\frac{19}{9} \\ & C=\frac{6}{9}-\frac{3 \cdot 19}{9}=\frac{2-19}{3}=-\frac{17}{3} \\ & A =-\frac{1}{9}\end{aligned}\right. \\ \\ \\ \int {(-\frac{1}{9x} + \frac{19}{9 \cdot (x+3)} - \frac{17}{3 \cdot (x+3)^2}}}) \, dx= -\frac{1}{9} \int \frac{dx}{x} + \frac{19}{9} \int \frac{dx}{x+3} - \frac{17}{3} \int \frac{dx}{x+3)^2} =$

$\\\\ =-\frac{1}{9} \ln{|x|} + \frac{19}{9} \ln{|x+3|} - \frac{17}{3} \cdot (- \frac{1}{x+3}) + C = \\ \\ = -\frac{1}{9} \ln{|x|} + \frac{19}{9} \ln{|x+3|} + \frac{17}{ 3 \cdot (x+3)}+ C$