The value of the integral $I = \int_0^1 \frac{1}{\sqrt{x^2 + 2x + 3}} dx$ is:

The integral is $I = \int_0^1 \frac{1}{\sqrt{x^2 + 2x + 3}} dx$

Complete the square for $x^2 + 2x + 3$:

$x^2 + 2x + 3$

$= (x + 1)^2 - 1 + 3$

$= (x + 1)^2 + 2$

The integral becomes:

$I = \int_0^1 \frac{1}{\sqrt{(x+1)^2 + 2}} dx$

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Let $u = x + 1$

Then $du = dx$

When $x = 0$: $u = 1$

When $x = 1$: $u = 2$

$I = \int_1^2 \frac{1}{\sqrt{u^2 + 2}} du$

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Using the standard integral $\int \frac{1}{\sqrt{u^2 + a^2}} du = \ln|u + \sqrt{u^2 + a^2}| + C$ with $a^2 = 2$:

$I = \left[\ln\left|u + \sqrt{u^2 + 2}\right|\right]_1^2$

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At $u = 2$:

$\ln\left|2 + \sqrt{4 + 2}\right|$

$= \ln(2 + \sqrt{6})$

At $u = 1$:

$\ln\left|1 + \sqrt{1 + 2}\right|$

$= \ln(1 + \sqrt{3})$

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

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

Therefore, the value of the integral is $\ln\left(\frac{2 + \sqrt{6}}{1 + \sqrt{3}}\right)$