Let \(u = e^x\). Then \(e^{-x} = \frac{1}{u}\).

The equation becomes:

\(\frac{u + \frac{1}{u}}{u + 1} = 4\).

Multiply through by \(u(u + 1)\) to clear the fraction:

\(u^2 + 1 = 4u(u + 1)\).

Expand and simplify:

\(u^2 + 1 = 4u^2 + 4u\).

Rearrange to form a quadratic equation:

\(3u^2 + 4u - 1 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 3\), \(b = 4\), \(c = -1\):

\(u = \frac{-4 \pm \sqrt{4^2 - 4 \times 3 \times (-1)}}{2 \times 3}\).

\(u = \frac{-4 \pm \sqrt{16 + 12}}{6}\).

\(u = \frac{-4 \pm \sqrt{28}}{6}\).

\(u = \frac{-4 \pm 2\sqrt{7}}{6}\).

\(u = \frac{-2 \pm \sqrt{7}}{3}\).

Choose the positive root: \(u = \frac{\sqrt{7} - 2}{3}\).

Since \(u = e^x\), solve for \(x\):

\(x = \ln\left(\frac{\sqrt{7} - 2}{3}\right)\).

Calculate \(x\) to 3 decimal places:

\(x \approx -1.536\).