Notice that:
[tex]1^2-1=0.[/tex]
Therefore:
[tex]f(x)=\frac{x-1}{x^2-1}[/tex]
is undefined at x=1.
Now, recall that the natural logarithm is defined over the positive numbers, therefore:
[tex]x^2-1>0[/tex]
Then:
[tex]x>1\text{ or }x<-1.[/tex]
Therefore the function
[tex]f(x)=ln(x^2-1)[/tex]
is not well defined over the interval (0,2).
Finally, notice that:
[tex]p(x)=x^2+1,[/tex]
has no real zeros, therefore the function:
[tex]f(x)=\frac{x+1}{x^2+1}[/tex]
is well defined over all real numbers, also is continuous over all real numbers, particularly over the interval (0,2).
Answer: First option.
II only.
