How do i integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$? Integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$
I put $x^2+3x+2=t,$ and notice that $2x+3 dx=dt$, but the $dx$ is above! Please help me!

Hint
For $(x+2)(x+1)>0$
We need $x>-1$ or $x<-2$
$2x<-4$ or $>-2$
$2x+3<-1$ or $>1$
WLOG $2x+3=\sec t,-\dfrac\pi2<t<\dfrac\pi2$
$2dx=\sec t\tan t\ dt$
$x^2+3x+2=\dfrac{(2x+3)^2-1}4=\dfrac{\tan^2t}4$
