The value of the integral I = ∫limits_0^1 1/(√(x^2 + 2x + 3)) dx is:

The correct answer is Option (2) → \( \log\left( \frac{2 + \sqrt{6}}{1 + \sqrt{3}} \right) \)

Compute $I=\int_{0}^{1} \frac{dx}{\sqrt{x^2+2x+3}}$.

Complete the square: $x^2+2x+3=(x+1)^2+2$.

Let $u=x+1 \Rightarrow du=dx$. When $x=0$, $u=1$; when $x=1$, $u=2$.

Thus $I=\int_{1}^{2} \frac{du}{\sqrt{u^2+2}}=\int_{1}^{2} \frac{du}{\sqrt{u^2+(\sqrt{2})^2}}$.

Standard form: $\displaystyle \int \frac{du}{\sqrt{u^2+a^2}}=\ln\!\left|u+\sqrt{u^2+a^2}\right|+C$.

Apply with $a=\sqrt{2}$:

$I=\left[\ln\!\left(u+\sqrt{u^2+2}\right)\right]_{1}^{2}$

$I=\ln\!\left(2+\sqrt{4+2}\right)-\ln\!\left(1+\sqrt{1+2}\right)$

$I=\ln\!\left(2+\sqrt{6}\right)-\ln\!\left(1+\sqrt{3}\right)=\ln\!\left(\frac{2+\sqrt{6}}{1+\sqrt{3}}\right)$

Result: $\displaystyle I=\ln\!\left(\frac{2+\sqrt{6}}{1+\sqrt{3}}\right)$