Integrate the function: $\frac{x+2}{\sqrt{x^{2}+2 x+3}}$

We need to evaluate the integral $I = \int \frac{x+2}{\sqrt{x^{2}+2 x+3}} dx$.
First,we express the numerator in terms of the derivative of the quadratic expression inside the square root,which is $\frac{d}{dx}(x^2+2x+3) = 2x+2$.
$\int \frac{x+2}{\sqrt{x^{2}+2 x+3}} dx = \frac{1}{2} \int \frac{2(x+2)}{\sqrt{x^{2}+2 x+3}} dx = \frac{1}{2} \int \frac{2x+4}{\sqrt{x^{2}+2 x+3}} dx$
$= \frac{1}{2} \int \frac{2x+2+2}{\sqrt{x^{2}+2 x+3}} dx = \frac{1}{2} \int \frac{2x+2}{\sqrt{x^{2}+2 x+3}} dx + \int \frac{1}{\sqrt{x^{2}+2 x+3}} dx$
Let $I_1 = \int \frac{2x+2}{\sqrt{x^{2}+2 x+3}} dx$ and $I_2 = \int \frac{1}{\sqrt{x^{2}+2 x+3}} dx$.
For $I_1$,let $t = x^2+2x+3$,then $dt = (2x+2)dx$.
$I_1 = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{x^2+2x+3}$.
For $I_2$,complete the square: $x^2+2x+3 = (x+1)^2 + 2 = (x+1)^2 + (\sqrt{2})^2$.
$I_2 = \int \frac{dx}{\sqrt{(x+1)^2 + (\sqrt{2})^2}} = \log |(x+1) + \sqrt{(x+1)^2 + 2}| = \log |(x+1) + \sqrt{x^2+2x+3}|$.
Combining these,$I = \frac{1}{2}(2\sqrt{x^2+2x+3}) + \log |(x+1) + \sqrt{x^2+2x+3}| + C$.
Final result: $I = \sqrt{x^2+2x+3} + \log |(x+1) + \sqrt{x^2+2x+3}| + C$.