<s>[INST] 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!
 [/INST] 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)
$I = 2.\frac{1}{2}.tan^{-1}(2t)+ c = tan^{-1}(2t)+c= tan^{-1}(2\sqrt{x^2+3x+2})+c = sec^{-1}(2x+3) +c = cos^{-1}(\frac{1}{2x+3})+c = \pi/2 - sin^{-1}(\frac{1}{2x+3}) +c = -sin^{-1}(\frac{1}{2x+3}) +c'$
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