Answer
$$\frac{1}{2}[(x-2) \sqrt{x^{2}-4 x+3}-\ln |\sqrt{x^{2}-4 x+3}+x-2|]+C$$
Work Step by Step
Given $$\int \sqrt{x^{2}-4 x+3} d x$$
Since
\begin{align*}
x^{2}-4 x+3&=(x-2)^{2}-4 +3\\
&=(x-2)^{2}-1
\end{align*}
Let $$x-2= \sec u\ \ \ \ \ \ dx=\sec u\tan u du $$
then
\begin{align*}
\int \sqrt{x^{2}-4 x+3} d x&=\int \sqrt{(x-2)^{2}-1} d x\\
&=\int \sqrt{ \sec^2 u-1} \sec u\tan u du\\
&= \int \tan^2 u\sec udu \\
&= \int (\sec^3 u-\sec u)du
\end{align*}
Use
$$\int \sec ^{3} u d u=\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C $$
Hence
\begin{align*}
\int \sqrt{x^{2}-4 x+3} d x&=\int \sqrt{(x-2)^{2}-1} d x\\
&=\int \sqrt{ \sec^2 u-1} \sec u\tan u du\\
&= \int \tan^2 u\sec udu \\
&= \int (\sec^3 u-\sec u)du \\
&= \frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+\ln|\sec u+\tan u| +C\\
&= \frac{1}{2} \sec u \tan u-\frac{1}{2} \ln |\sec u+\tan u| +C\\
&= \frac{1}{2}[(x-2) \sqrt{x^{2}-4 x+3}-\ln |\sqrt{x^{2}-4 x+3}+x-2|]+C
\end{align*}
