Answer: \(\ln(2x+1)-\ln(2x-1)\); \(\dfrac{dy}{dx}=\dfrac{16x^2-8}{4x^2-1}\); stationary point \(x=\dfrac1{\sqrt2}\), a minimum.

Using logarithm laws,

\(\ln\left(\dfrac{2x+1}{2x-1}\right)=\ln(2x+1)-\ln(2x-1).\)

So

\(y=\ln(2x+1)-\ln(2x-1)+4x.\)

Differentiate:

\(\dfrac{dy}{dx}=\dfrac{2}{2x+1}-\dfrac{2}{2x-1}+4.\)

Put the terms over the denominator \(4x^2-1=(2x+1)(2x-1)\):

\(\dfrac{dy}{dx}=\dfrac{2(2x-1)-2(2x+1)+4(4x^2-1)}{4x^2-1}.\)

The numerator simplifies to

\(4x-2-4x-2+16x^2-4=16x^2-8.\)

Hence

\(\dfrac{dy}{dx}=\dfrac{16x^2-8}{4x^2-1}.\)

For a stationary point, set \(\dfrac{dy}{dx}=0\):

\(16x^2-8=0.\)

Thus

\(x^2=\dfrac12.\)

Since the curve is defined for \(x\gt \dfrac12\), the valid solution is

\(x=\dfrac1{\sqrt2}.\)

For \(x\gt \dfrac12\), the denominator \(4x^2-1\) is positive. The numerator \(16x^2-8\) changes from negative to positive at \(x=\dfrac1{\sqrt2}\).

So the stationary point is a minimum.