Answer: \(y=-\dfrac{4+\sqrt2}{12}(x-\sqrt2)\); approximate change \(\dfrac{24-6\sqrt2}{7}h\).

Let

\(y=\frac{\ln(3x^2-5)}{2x+1}.\)

Using the quotient rule,

\(\frac{dy}{dx} =\frac{(2x+1)\dfrac{6x}{3x^2-5}-2\ln(3x^2-5)}{(2x+1)^2}.\)

When \(x=\sqrt2\),

\(3x^2-5=1,\)

so \(\ln(3x^2-5)=\ln1=0\), and the point on the curve is

\((\sqrt2,0).\)

The gradient of the tangent is

\(\frac{dy}{dx} =\frac{(2\sqrt2+1)6\sqrt2}{(2\sqrt2+1)^2} =\frac{6\sqrt2}{2\sqrt2+1}.\)

Rationalising,

\(\frac{6\sqrt2}{2\sqrt2+1} =\frac{24-6\sqrt2}{7}.\)

The gradient of the normal is the negative reciprocal:

\(-\frac{2\sqrt2+1}{6\sqrt2} =-\frac{4+\sqrt2}{12}.\)

Hence the equation of the normal is

\(y=-\frac{4+\sqrt2}{12}(x-\sqrt2).\)

For a small increase \(h\), the approximate change in \(y\) is

\(\Delta y\approx \frac{dy}{dx}h.\)

Therefore

\(\Delta y\approx \frac{24-6\sqrt2}{7}h.\)