Answer
$$\frac{3}{2}\left[\frac{(x-2) \sqrt{x^{2}-4 x+7}}{3}+\ln \left|\frac{(x-2)+\sqrt{x^{2}-4 x+7}}{\sqrt{3}}\right|\right]+C$$
Work Step by Step
Given $$\int \sqrt{x^{2}-4 x+7} d x$$
Since
\begin{align*}
x^{2}-4 x+7&=(x-2)^{2}-4 +7\\
&=(x-2)^{2}+3
\end{align*}
Let $$x-2=\sqrt{3}\tan u\ \ \ \ \ \ dx=\sqrt{3}\sec^2 udu $$
then
\begin{align*}
\int \sqrt{x^{2}-4 x+7} d x&=\int \sqrt{(x-2)^{2}+3} d x\\
&=\sqrt{3}\int \sqrt{3\tan^{2}u+3} \sec^2 udu \\\\
&=3\int \sec^3udu
\end{align*}
Use
$$\int \sec ^{n} u d u=\frac{1}{n-1} \tan u \sec ^{n-2} u+\frac{n-2}{n-1} \int \sec ^{n-2} u d u $$
then
$$\int \sec ^{3} u d u=\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C$$
Hence
\begin{align*}
\int \sqrt{x^{2}-4 x+7} d x &=3\int \sec^3udu\\
&= \frac{3}{2} \sec u \tan u+\frac{3}{2} \ln |\sec u+\tan u|+C\\
&=\frac{3}{2}\left[\frac{u \sqrt{u^{2}+3}}{3}+\ln \left|\frac{u+\sqrt{u^{2}+3}}{\sqrt{3}}\right|\right]+C\\
&=\frac{3}{2}\left[\frac{(x-2) \sqrt{x^{2}-4 x+7}}{3}+\ln \left|\frac{(x-2)+\sqrt{x^{2}-4 x+7}}{\sqrt{3}}\right|\right]+C
\end{align*}
