Posted by strawberryfields on Friday, January 23, 2009 at 10:18pm.
ln[x/(x^2-1)^2]
would you mind telling me how you got it?
i don't get this problem..
thanks forthe answer though!
Actually I should have written it as
ln{[x/(x^2-1]^2}
The -ln(x+1)-ln(x-1) in the denominator is ln {1/[(x+1)(x-1)] = ln [1/(x^2-1)]
Adding ln x puts the x in the numerator of [x/(x^2-1)]
The two in front of
2[ln(x)-ln(x+1)-ln(x-1)]
is the same as taking the log of the square of [ln(x)-ln(x+1)-ln(x-1)]
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