Question 367894
{{{drawing(300,300,-2,4,-3,3,circle(1.384019,0,0.2),graph(300,300,-2,4,-3,3,0,(3x^2-4x)/(x+1)+ln(x^2-1)))}}}

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I used Newton's method with a starting value of {{{x[1]=1}}}.
{{{f=(3x^2-4x)/(x+1)+ln(x^2-1)))}}}
{{{df/dx=(3x^3+5x^2-8x+4)/((1+x)^2(x-1))}}}
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It quickly iterated to {{{x=1.384019}}}