ПОМОГИТЕ!НАЙДИТЕ ИНТЕГРАЛ!

Решите задачу:

$\int \frac{(x+2)dx}{(x-2)(x^2+2x+4)} =I\\\\ \frac{x+2}{(x-2)(x^2+2x+4)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+2x+4} = \frac{A(x^2+2x+4)+(Bx+C)(x-2)}{(x-2)(x^2+2x+4)} \\\\x+2=A(x^2+2x+4)+(Bx+C)(x-2)\\\\x+2=(A+B)x^2+(2A-2B+C)x+(4A-2C)x^0\; ,\; \; [x^0=1]\\\\x^2\; |\, A+B=0,\; \; \to \; \; A=-B\\\\x\; |\; 2A-2B+C=1,\; \; \to \; \; C=1+4B,\\\\x^0\; |\; 4A-2C=2,\; \; \to \; \; -2B-C=1,\; C=-2B-1\\\\-2B-1=1+4B\; \; \to \; \; 6B=-2,\; B=-\frac{1}{3}\\\\A=\frac{1}{3},\; \; C=-\frac{1}{3}$

$I=\frac{1}{3}\int \frac{dx}{x-2}-\frac{1}{3}\int \frac{x+1}{x^2+2x+4}dx=\frac{1}{3}ln|x-2|-\frac{1}{3}\int \frac{x+1}{(x+1)^2+3} =\\\\=[\, t=x+1,\; x=t-1,\; dt=dx\, ]=\frac{1}{3}ln|x-2|-\frac{1}{3}\int \frac{t\, dt}{t^2+3} =\\\\=\frac{1}{3}ln|x-2|-\frac{1}{3}\cdot \frac{1}{2}\int \frac{2t\, dt}{t^2+3}=\frac{1}{3}ln|x-2|-\frac{1}{6}\int \frac{du}{u}=[\, u=t^2+3\, ]=\\\\=\frac{1}{3}ln|x-2|-\frac{1}{6}ln|u|+C=\frac{1}{3}ln|x-2|-\frac{1}{6}ln|x^2+2x+4|+C$