Интеграл xdx/x^2+2x+4

$\int{ \frac{x}{x^2+2x+4} } \, dx =\int{ \frac{x}{x^2+2x+1+3} } \, dx =\int{ \frac{x}{(x+1)^2+3} } \, dx$

Замена
х+1=t
x=t-1
dx=dt
$\int{ \frac{t-1}{t^2+3} } \, dt=\int{ \frac{t}{t^2+3} } \, dt-\int{ \frac{1}{t^2+3} } \, dt= \frac{1}{2} ln|t^2+3|- \frac{1}{ \sqrt{3} } arctg \frac{t}{ \sqrt{3} }+C= \\ \\ = \frac{1}{2} ln|x^2+2x+4|- \frac{1}{ \sqrt{3} } arctg \frac{x+1}{ \sqrt{3} }+C$