Integrate $\int \frac{\mathrm dx}{(x-2)(\sqrt{x^2-4x+3})}$

Question

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 $-\arctan\left(\frac{1}{\sqrt{x^2-4x+3}}\right)+\rm C\;.$

Answer 1

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

Answer 2

$$\begin{align} \int\frac{\mathrm{d}x}{\left(x-2\right)\sqrt{x^{2}-4x+3}} &=\int\frac{\mathrm{d}x}{\left(x-2\right)\sqrt{\left(x-1\right)\left(x-3\right)}}\\ &=\int\frac{\mathrm{d}t}{t\sqrt{t^{2}-1}};~~~\small{\left[x=t+2\right]}\\ &=\int\frac{1}{t^{2}}\cdot\frac{t\,\mathrm{d}t}{\sqrt{t^{2}-1}}\\ &=\int\frac{\mathrm{d}u}{1+u^{2}};~~~\small{\left[\sqrt{t^{2}-1}=u\right]}\\ &=-\int\frac{\left(-u^{-2}\right)\,\mathrm{d}u}{u^{-2}+1}\\ &=-\int\frac{\mathrm{d}w}{w^{2}+1};~~~\small{\left[u^{-1}=w\right]}\\ &=-\arctan{\left(w\right)}+\color{grey}{constant}\\ &=-\arctan{\left(\frac{1}{u}\right)}+\color{grey}{constant}\\ &=-\arctan{\left(\frac{1}{\sqrt{t^{2}-1}}\right)}+\color{grey}{constant}\\ &=-\arctan{\left(\frac{1}{\sqrt{\left(x-2\right)^{2}-1}}\right)}+\color{grey}{constant}.\blacksquare\\ \end{align}$$