Lời giải:
\(y'=\ln (x+1+\sqrt{x^2+2x+3})'=\frac{(x+1+\sqrt{x^2+2x+3})'}{x+1+\sqrt{x^2+2x+3}}=\frac{1+\frac{2x+2}{2\sqrt{x^2+2x+3}}}{x+1+\sqrt{x^2+2x+3}}\)
\(=\frac{\sqrt{x^2+2x+3}+x+1}{(x+1+\sqrt{x^2+2x+3})\sqrt{x^2+2x+3}}=\frac{1}{\sqrt{x^2+2x+3}}\)