question_answer

\[\int{\frac{{{e}^{x}}-1}{{{e}^{x}}+a}}\,\,dx\] is equal to _____.

A)  \[\log \left( {{e}^{x}}+2\sqrt{{{e}^{2x}}-1} \right)-{{\sec }^{-1}}{{e}^{2x}}+C\]

B)  \[\log \left( {{e}^{x}}+\sqrt{{{e}^{2x}}-1} \right)-{{\sec }^{-1}}\left( {{e}^{x}} \right)+C\]

C)  \[\log \left( {{e}^{x}}-\sqrt{{{e}^{2x}}-1} \right)-{{\sec }^{-1}}{{e}^{x}}+C\]

D)  \[\log \left( {{e}^{x}}-\sqrt{{{e}^{2x}}-1} \right)+{{\sec }^{-1}}{{e}^{x}}+C\]

E)  None of these