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