1. The problem asks to find the integral of the function $P(u) = \frac{4n-1}{x-1}$.\n\n2. The integral formula for a function of the form $\frac{a}{x-b}$ is $$\int \frac{a}{x-b} \, dx = a \ln|x-b| + C$$ where $C$ is the constant of integration.\n\n3. Applying this formula to $P(u)$, we treat $4n-1$ as a constant with respect to $x$, so:\n$$\int \frac{4n-1}{x-1} \, dx = (4n-1) \int \frac{1}{x-1} \, dx = (4n-1) \ln|x-1| + C$$\n\n4. Therefore, the integral of $P(u)$ is $$\boxed{(4n-1) \ln|x-1| + C}$$ where $C$ is the constant of integration.