The value of the integral \( \displaystyle \int\limits_{-1}^{1}\log_{e}\!\left(\sqrt{1-x}+\sqrt{1+x}\right)\,dx \) is equal to:
\( 2\log_{e}2+\dfrac{\pi}{2}-\dfrac{1}{2} \)
\( \dfrac{1}{2}\log_{e}2+\dfrac{\pi}{4}-\dfrac{3}{2} \)
\( \log_{e}2+\dfrac{\pi}{2}-1 \)
\( 2\log_{e}2+\dfrac{\pi}{4}-1 \)