ПОМОГИТЕ!НАЙДИТЕ ИНТЕГРАЛ!

Решите задачу:

$\int \frac{x^2-5x+1}{(x-1)(x^2+2x+4)} dx=I\\\\ \frac{x^2-5x+1}{(x-1)(x^2+2x+4)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+2x+4} = \frac{A(x^2+2x+4)+(Bx+C)(x-1)}{(x-1)(x^2+2x+4)} \\\\x^2-5x+1=A(x^2+2x+4)+(Bx+C)(x-1)\\\\x^2\; |\; A+B=1,\; \; \to \; \; B=1-A\\\\x\; |\; 2A-B+C=-5,\; \; \to \; \; 3A+C=-4\\\\x^0\; |\; 4A-C=1,\; \; \to \; \; C=4A-1\; ,\; \\\\3A+C=3A+4A-1=-4,\; \; 7A=-3,\; A=-\frac{3}{7} \\\\C=-\frac{12}{7}-1=-\frac{19}{7} ,\; B=1+\frac{3}{7}=\frac{10}{7}$

$I=-\frac{3}{7}\int \frac{dx}{x-1}+\int \frac{\frac{10}{7}x-\frac{19}{7}}{x^2+2x+4} dx=-\frac{3}{7}ln|x-1|+\int \frac{\frac{10}{7}x-\frac{19}{7}}{(x+1)^2+3} dx=\\\\=[t=x+1,\; x=t-1,\; dx=dt]=-\frac{3}{7}ln|x-1|+\int \frac{\frac{10}{7}t-\frac{29}{7}}{t^2+3}dt=\\\\=-\frac{3}{7}ln|x-1|+\frac{10}{7}\cdot \frac{1}{2}\int \frac{2t\, dt}{t^2+3}-\frac{29}{7}\int \frac{dt}{t^2+3}=-\frac{3}{7}ln|x-1|+\\\\+\frac{5}{7}ln|t^2+3|-\frac{29}{7}\cdot \frac{1}{\sqrt3}arctg\frac{t}{\sqrt3}+C=-\frac{3}{7}ln|x-1|+$

$+\frac{5}{7}ln|x^2+2x+4|-\frac{29}{7\sqrt3}arctg\frac{x+1}{\sqrt3}+C$