<s>[INST] Integrate $\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}$ 
Integrate $$\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}$$

$$\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}=\int \frac{\mathrm dx}{(x-2)(\sqrt{(x-2)^2-(1)^2})}=\operatorname{arcsec}(x-2)+\rm C$$
Wolfram writes that the answer is $-\arctan\left(\frac{1}{\sqrt{x^2-4x+3}}\right)+\rm C\;.$
 [/INST] if you draw a right angled triangle you can see that $\operatorname{arcsec}(x-2)$ is equivalent to $\arctan\sqrt{x^2-4x+3}$ which in turn is equivalent to $\frac{\pi}{2}-\arctan\frac{1}{\sqrt{x^2-4x+3}}$, so the answers are the same give or take a constant
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