The value of the integral I = ∫_0^1 1/(√(x^2 + 2x + 3)) dx is:

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

Completing the square in the denominator:

    \(x^2 + 2x + 3 = (x + 1)^2 + 2\)

Let u = x + 1, so du = dx; the limits x = 0 and x = 1 become u = 1 and u = 2:

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

By the standard integral \(\int \frac{du}{\sqrt{u^2 + a^2}} = \log\left|u + \sqrt{u^2 + a^2}\right| + C\):

    \( \displaystyle I = \Big[\log\left(u + \sqrt{u^2 + 2}\right)\Big]_{1}^{2} \)

    \( \displaystyle I = \log\left(2 + \sqrt{6}\right) - \log\left(1 + \sqrt{3}\right) \)

    \( \displaystyle I = \log\left(\frac{2 + \sqrt{6}}{1 + \sqrt{3}}\right) \)