College Algebra Ln(log) problem how do you write it as a single logarithm

I have been trying to figure out this problem and can't if you could show me how to write it as a single logarithm, I would be much obliged, also please show me how you did it step by step so i can figure out how to do the rest of my homework, Thanks!

57.) ln(x/x-1)+ln(x+1/x)-ln(x²-1)

Thanks so much!!!

ln(x / (x - 1)) + ln((x + 1) / x) - ln(x^2 - 1) =>
ln(x) - ln(x - 1) + ln(x + 1) - ln(x) - ln((x - 1) * (x + 1)) =>
ln(x) - ln(x) - ln(x - 1) + ln(x + 1) - ln(x - 1) - ln(x + 1) =>
0 - 2 * ln(x - 1) + ln(x - 1) - ln(x + 1) =>
-2 * ln(x - 1)

Remember:

ln(a / b) = ln(a) - ln(b)
ln(a * b) = ln(a) + ln(b)

I assume that x/x-1 should be x/(x-1) and x+1/x should be (x+1)/x

ln[x/(x-1)] + ln[(x+1)/x] - ln(x²-1)
= ln[x/(x-1) * (x+1)/x] - ln(x²-1)
= ln[(x+1)/(x-1)] - ln[(x+1)(x-1)]
= ln{(x+1)/(x-1) * 1/[(x+1)(x-1)]}
= ln[1/(x-1)²]
= -ln[(x-1)²]
= -2ln(x-1)