Đặt $t=x+2$
$\to dt=dx$
Có $t=x+2\to x=t-2\to x^2=(t-2)^2$
$I=\displaystyle\int \dfrac{1-(t-2)^2}{t}dt$
$=\displaystyle\int \dfrac{1-(t^2-4t+4)}{t}dt$
$=\displaystyle\int \dfrac{-t^2+4t-3}{t}dt$
$=\displaystyle\int \left( -t+4-\dfrac{3}{t}\right)dt$
$=\dfrac{-1}{2}t^2+4t-3\ln |t|+C$
$=\dfrac{-1}{2}(x+2)^2+4(x+2)-3\ln|x+2|+C$
$=\dfrac{-1}{2}x^2+2x-3\ln|x+2|+C$