Интеграл dx/x(x^2+1)

Решите задачу:

$\int \frac{dx}{x(x^2+1)}=\int(\frac{A}x}+\frac{Bx+C}{x^2+1})dx=I\\\\\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+x(Bx+C)}{x(x^2+1)}\\\\1=Ax^2+Bx^2+Cx+A\\\\x^2|\; A+B=0\\\\x\; |\; C=0\\\\x^0|\; A=1\\\\B=-A=-1\\\\I=\int \frac{dx}{x}+\int \frac{-x\, dx}{x^2+1}=ln|x|-\frac{1}{2}\int \frac{2x\, dx}{x^2+1}=ln|x|-\frac{1}{2}\cdot ln(x^2+1)+C$