<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] 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, $I=\int \frac{dz}{z\sqrt{z^2-1}}=sec^{-1}z=sec^{-1}(2x+3)$
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