The value of the definite integral $I = \int_{-1}^{1} \frac{1}{1 + \sqrt{e^x}} dx$ is:

The integral to evaluate is:

$I = \int_{-1}^{1} \frac{1}{1 + \sqrt{e^x}} dx$

This integral has symmetric limits from $-1$ to $1$. For such integrals, the following property can be used:

$I = \int_{-1}^{1} f(x) dx = \int_{-1}^{1} f(-x) dx$

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Given $f(x) = \frac{1}{1 + \sqrt{e^x}}$, substitute $-x$:

$f(-x) = \frac{1}{1 + \sqrt{e^{-x}}}$

Since $e^{-x} = \frac{1}{e^x}$:

$f(-x) = \frac{1}{1 + \frac{1}{\sqrt{e^x}}}$

Multiplying numerator and denominator by $\sqrt{e^x}$:

$f(-x) = \frac{\sqrt{e^x}}{\sqrt{e^x} + 1}$

$f(-x) = \frac{\sqrt{e^x}}{1 + \sqrt{e^x}}$

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Adding $f(x)$ and $f(-x)$:

$f(x) + f(-x) = \frac{1}{1 + \sqrt{e^x}} + \frac{\sqrt{e^x}}{1 + \sqrt{e^x}}$

$f(x) + f(-x) = \frac{1 + \sqrt{e^x}}{1 + \sqrt{e^x}}$

$f(x) + f(-x) = 1$

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Since $I = \int_{-1}^{1} f(x) dx$ and $I = \int_{-1}^{1} f(-x) dx$:

$2I = \int_{-1}^{1} f(x) dx + \int_{-1}^{1} f(-x) dx$

$2I = \int_{-1}^{1} [f(x) + f(-x)] dx$

$2I = \int_{-1}^{1} 1 \, dx$

$2I = [x]_{-1}^{1}$

$2I = 1 - (-1)$

$2I = 2$

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Therefore:

$I = 1$