To solve the equation

\[ \frac{x+1}{x-1} + \frac{x-1}{x+1} = \frac{4}{(x+1)(x-1)}, \]

we first need to find a common denominator for the fractions on the left-hand side. The common denominator is \((x-1)(x+1)\).

Rewrite each fraction with the common denominator:

\[ \frac{(x+1)^2}{(x-1)(x+1)} + \frac{(x-1)^2}{(x-1)(x+1)} = \frac{4}{(x+1)(x-1)}. \]

Now, combine the fractions on the left-hand side:

\[ \frac{(x+1)^2 + (x-1)^2}{(x-1)(x+1)} = \frac{4}{(x+1)(x-1)}. \]

Next, expand the numerators:

\[ (x+1)^2 = x^2 + 2x + 1, \]
\[ (x-1)^2 = x^2 - 2x + 1. \]

So,

\[ (x+1)^2 + (x-1)^2 = x^2 + 2x + 1 + x^2 - 2x + 1 = 2x^2 + 2. \]

Thus, the equation becomes:

\[ \frac{2x^2 + 2}{(x-1)(x+1)} = \frac{4}{(x+1)(x-1)}. \]

Since the denominators are the same, we can equate the numerators:

\[ 2x^2 + 2 = 4. \]

Subtract 2 from both sides:

\[ 2x^2 = 2. \]

Divide by 2:

\[ x^2 = 1. \]

Take the square root of both sides:

\[ x = \pm 1. \]

However, we need to check if these solutions are valid by substituting them back into the original equation.

For \( x = 1 \):

\[ \frac{1+1}{1-1} + \frac{1-1}{1+1} = \frac{4}{(1+1)(1-1)} \]

This results in division by zero, which is undefined. So, \( x = 1 \) is not a valid solution.

For \( x = -1 \):

\[ \frac{-1+1}{-1-1} + \frac{-1-1}{-1+1} = \frac{4}{(-1+1)(-1-1)} \]

This also results in division by zero, which is undefined. So, \( x = -1 \) is not a valid solution.

Therefore, the equation

\[ \frac{x+1}{x-1} + \frac{x-1}{x+1} = \frac{4}{(x+1)(x-1)} \]

has no valid solutions.