Let $I=\int \frac{x+2}{\sqrt{x^2+2x+3}}dx$ $=\frac{1}{2}\int \frac{2x+4}{\sqrt{x^2+2x+3}}dx$ $=\frac{1}{2}\int \frac{2x+2+2}{\sqrt{x^2+2x+3}}dx$ $=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\int \frac{dx}{\sqrt{x^2+2x+3}}$ Let $I = I_1 + I_2$ ...(i) where $I_1=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx$ Let $x^2 + 2x + 3 = t$ $\Rightarrow (2x + 2) dx = dt$ $\therefore I_1 = \frac{1}{2}\int \frac{dt}{\sqrt{t}} = \frac{1}{2} \times 2 t^{1/2}$ $=\frac{1}{2} \times 2 \sqrt{x^2 + 2x + 3}+C_1$ $=\sqrt{x^2 + 2x + 3}+C_1$ ...(ii) Also, $I_2=\int \frac{dx}{\sqrt{x^2+2x+3}}$ $=\int \frac{dx}{\sqrt{x^2+2x+1-1+3}}$ $=\int \frac{dx}{\sqrt{(x+1)^2+(\sqrt{2})^2}}$ $=\log \left| (x+1)+\sqrt{(x+1)^2+2} \right|$ $=\log \left| (x+1)+\sqrt{x^2+2x+3} \right|+C_2$ ... (iii) Hence from (i), (ii) and (iii), we get