Simplify \( \frac{4 x}{x-1}+\frac{2}{(x-1)^{2}} \) define the domain \( x \neq \square \) State the sum in simplest form.

To simplify \( \frac{4 x}{x-1}+\frac{2}{(x-1)^{2}} \), we need a common denominator, which is \((x-1)^{2}\). Rewriting the first term: \[ \frac{4x}{x-1} = \frac{4x(x-1)}{(x-1)(x-1)} = \frac{4x^2 - 4x}{(x-1)^{2}} \] Now we can combine the fractions: \[ \frac{4x^2 - 4x + 2}{(x-1)^{2}} \] This can be simplified further by combining like terms in the numerator: \[ 4x^2 - 4x + 2 \] Thus, the expression simplifies to: \[ \frac{4x^2 - 4x + 2}{(x-1)^{2}} \] Now, let’s define the domain. The denominator \((x-1)^{2}\) cannot be zero, so we set \(x-1 \neq 0\), which gives us \(x \neq 1\). Therefore, the solution can be stated as: The domain is \(x \neq 1\), and the simplest form is \[ \frac{4x^2 - 4x + 2}{(x-1)^{2}}. \]