we can try going with IBP: #int u v' = uv - int u'v#
so
#int \ ln(x^2 -1) \ dx#
#= int \d/dx(x) ln(x^2 -1) \ dx#
and by IBP
#= xln(x^2 -1) - int \xd/dx (ln(x^2 -1)) \ dx#
#= xln(x^2 -1) - int \x 1/(x^2 -1)* 2x \ dx#
#= xln(x^2 -1) - 2 int \x^2/(x^2 -1) \ dx#
#= xln(x^2 -1) - 2 int \ (x^2- 1 + 1)/(x^2 -1) \ dx#
#= xln(x^2 -1) - 2 int \ 1 + 1/(x^2 -1) \ dx#
#= xln(x^2 -1) - 2x - 2 int \ 1/(x^2 -1) \ dx#
#= xln(x^2 -1) - 2x - 2 int \ 1/((x -1)(x+1)) \ dx#
so now using partial fractions
# 1/((x -1)(x+1)) = 1/2( 1/(x-1) - 1/(x+1))#
#implies xln(x^2 -1) - 2x - int \ 1/(x-1) - 1/(x+1) \ dx#
#= xln(x^2 -1) - 2x - ln (x-1) + ln (x+1) + C#
#= xln(x^2 -1) - 2x + ln ((x+1)/(x-1)) + C#