Answer: \(y=0.923x-0.692\).

Let

\(y=\frac{\ln(3x^2-1)}{x+2}.\)

Using the quotient rule,

\(\frac{dy}{dx} =\frac{(x+2)\left(\frac{6x}{3x^2-1}\right)-\ln(3x^2-1)}{(x+2)^2}.\)

When \(x=1\),

\(y=\frac{\ln(3(1)^2-1)}{1+2} =\frac{\ln2}{3}.\)

The gradient at \(x=1\) is

\(\frac{dy}{dx} =\frac{3\left(\frac{6}{2}\right)-\ln2}{9} =\frac{9-\ln2}{9}.\)

So

\(m=1-\frac{\ln2}{9}=0.92298\ldots.\)

Hence \(m=0.923\) to 3 decimal places.

The tangent has equation

\(y-\frac{\ln2}{3}=m(x-1).\)

Thus

\(c=\frac{\ln2}{3}-m.\)

Using \(m=\frac{9-\ln2}{9}\),

\(c=\frac{\ln2}{3}-\frac{9-\ln2}{9} =\frac{4\ln2}{9}-1.\)

So

\(c=-0.6919\ldots,\)

and hence \(c=-0.692\) to 3 decimal places.

Therefore the tangent is

\(y=0.923x-0.692.\)