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

1

The correct answer is Option (2) → 1

Given definite integral: $I = \int_{-1}^{1} \frac{1}{1 + \sqrt{e^x}} \, dx$

Substitute $t = \sqrt{e^x} \Rightarrow t^2 = e^x, \; dx = \frac{2 dt}{t}$

Integral becomes:

$I = \int \frac{2}{t(1 + t)} \, dt = 2 \int \left( \frac{1}{t} - \frac{1}{1+t} \right) dt = 2 (\ln|t| - \ln|1+t|) + C = 2 \ln \left| \frac{t}{1+t} \right| + C$

Back-substitute $t = \sqrt{e^x}$:

$I = 2 \ln \left( \frac{e^{x/2}}{1 + e^{x/2}} \right)$

Evaluate from $x = -1$ to $x = 1$:

$I = 2 \ln \left( \frac{e^{1/2}}{1 + e^{1/2}} \right) - 2 \ln \left( \frac{e^{-1/2}}{1 + e^{-1/2}} \right)$

Simplify using $e^{-1/2} = 1/e^{1/2}$:

$I = 2 \ln \left( \frac{\sqrt{e}}{1 + \sqrt{e}} \cdot \frac{1 + 1/\sqrt{e}}{1/\sqrt{e}} \right) = 2 \ln \left( \frac{\sqrt{e}}{1 + \sqrt{e}} \cdot \frac{\sqrt{e} + 1}{1} \right)$

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