Answer
$$\int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+3}} =\sec ^{-1}(x-2)+c $$
Work Step by Step
Given $$\int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+3}}$$
So, we get
\begin{aligned} I&= \int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+3}}\\
&= \int \frac{d x}{(x-2) \sqrt{x^{2}-4 x+4-1}}\\
&=\int \frac{d x}{(x-2) \sqrt{(x-2)^{2}-1}} \\
\end{aligned}
Let $$\begin{array}{l}{x-2=\sec \theta} \\ {\Rightarrow d x=\sec \theta \tan \theta d \theta}\end{array}$$
So, we get
\begin{aligned} I&= \int \frac{\sec \theta \tan \theta \ \ d \theta}{\sec \theta \sqrt{\sec ^{2} \theta-1}}\\
&=\int \frac{\sec \theta \tan \theta \ \ d \theta}{\sec \theta \tan \theta} \\
& =\int d \theta\\
&=\theta+c\\
&=\sec ^{-1}(x-2)+c
\end{aligned}
