Answer
$$\ln(x^4-1)-\ln(x+1)$$
Work Step by Step
$\ln[(x^2+1)(x-1)] = \ln\left[\frac{(x^2+1)(x-1)(x+1)}{x+1}\right]= \ln\left[\frac{(x^2+1)(x^2-1)}{x+1}\right] = \ln\left[\frac{x^4-1}{x+1}\right]=\boxed{\ln(x^4-1)-\ln(x+1)}$
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