question_answer

\[\int{\frac{In\left( \frac{x-1}{x+1} \right)\,}{{{x}^{2}}-1}}\]dx is equal to

A)  \[\frac{1}{2}In{{\left( \frac{x-1}{x+1} \right)}^{2}}+C\]

B)  \[\frac{1}{2}In{{\left( \frac{x+1}{x-1} \right)}^{2}}+C\]

C)  \[\frac{1}{4}\left[ In{{\left( \frac{x-1}{x+1} \right)}^{2}} \right]+C\]      

D)  \[\frac{1}{4}\left[ In\left( \frac{x+1}{x-1} \right) \right]+C\]