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!
 A: Found a three line solution. Do not always be fooled by the quadratics lol.
If we let the original integral be $I$,note that $I=\int \frac{dx}{(2x+3)\sqrt{\frac{1}{4}(4x^2+12x+8})}=\int \frac{dx}{(2x+3)\frac{1}{2}\sqrt{(4x^2+12x+8})}=\int \frac{dx}{(2x+3)\frac{1}{2}\sqrt{(2x+3)^2-1}} $.
Now you know what...Put $2x+3=z \implies 2dx=dz \implies dx=\frac{dz}{2}$. Hence, $I=\int \frac{dz}{z\sqrt{z^2-1}}=sec^{-1}z=sec^{-1}(2x+3)$
A: Hint
For $(x+2)(x+1)>0$
We need $x>-1$ or $x<-2$
$2x<-4$ or $>-2$
$2x+3<-1$ or $>1$
WLOG $2x+3=\sec t,-\dfrac\pi2<t<\dfrac\pi2$
$2dx=\sec t\tan t\ dt$
$x^2+3x+2=\dfrac{(2x+3)^2-1}4=\dfrac{\tan^2t}4$
A: Let $ t = \sqrt{x^2+3x+2}$,  $dt = \frac{2x+3}{2\sqrt{x^2+ 3x +2}}dx$, $dx = \frac{2\sqrt{x^2+ 3x +2}}{2x+3}dt$
$I = \int{ \frac{2\sqrt{x^2+ 3x +2}}{2x+3}. \frac{1}{{(2x+3)}.{\sqrt{x^2+ 3x +2}}}dt} =\int{ \frac{2}{2x+3}. \frac{1}{{2x+3}}dt} =\int{ \frac{2}{4x^2+6x+9}dt} = 2\int{\frac{1}{4t^2 + 1}} $(on solving  x in terms of t)
$I = 2.\frac{1}{2}.tan^{-1}(2t)+ c = tan^{-1}(2t)+c= tan^{-1}(2\sqrt{x^2+3x+2})+c = sec^{-1}(2x+3) +c = cos^{-1}(\frac{1}{2x+3})+c = \pi/2 - sin^{-1}(\frac{1}{2x+3}) +c = -sin^{-1}(\frac{1}{2x+3}) +c'$
