We are given several math problems. We will solve problem IV, part b. It asks us to first show that $f(x) = -\frac{2-x}{(x-1)^2}$ can be written as $f(x) = -\frac{1}{(x-1)^2} + \frac{1}{x-1}$ and then to calculate $K = \int_{-1}^{0} f(x) dx$.

1. Problem Description

We are given several math problems. We will solve problem IV, part b. It asks us to first show that $f(x) = -\frac{2-x}{(x-1)^2}$ can be written as $f(x) = -\frac{1}{(x-1)^2} + \frac{1}{x-1}$ and then to calculate $K = \int_{-1}^{0} f(x) dx$.

2. Solution Steps

First, let's show that $f(x) = -\frac{2-x}{(x-1)^2} = -\frac{1}{(x-1)^2} + \frac{1}{x-1}$.

We can rewrite the right-hand side as a single fraction:

$-\frac{1}{(x-1)^2} + \frac{1}{x-1} = -\frac{1}{(x-1)^2} + \frac{x-1}{(x-1)^2} = \frac{-1 + x - 1}{(x-1)^2} = \frac{x-2}{(x-1)^2} = -\frac{2-x}{(x-1)^2}$. This is equal to the given $f(x)$.

Next, we want to find $K = \int_{-1}^{0} f(x) dx = \int_{-1}^{0} \left(-\frac{1}{(x-1)^2} + \frac{1}{x-1}\right) dx$.

We can split this into two integrals:

$K = \int_{-1}^{0} -\frac{1}{(x-1)^2} dx + \int_{-1}^{0} \frac{1}{x-1} dx$.

For the first integral, let $u = x-1$. Then $du = dx$. When $x = -1$, $u = -2$. When $x = 0$, $u = -1$.

$\int_{-1}^{0} -\frac{1}{(x-1)^2} dx = \int_{-2}^{-1} -\frac{1}{u^2} du = \int_{-2}^{-1} -u^{-2} du = \left[ u^{-1} \right]_{-2}^{-1} = \left[ \frac{1}{u} \right]_{-2}^{-1} = \frac{1}{-1} - \frac{1}{-2} = -1 + \frac{1}{2} = -\frac{1}{2}$.

For the second integral, let $u = x-1$. Then $du = dx$. When $x = -1$, $u = -2$. When $x = 0$, $u = -1$.

$\int_{-1}^{0} \frac{1}{x-1} dx = \int_{-2}^{-1} \frac{1}{u} du = \left[ \ln |u| \right]_{-2}^{-1} = \ln |-1| - \ln |-2| = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2)$.

Therefore, $K = -\frac{1}{2} - \ln(2)$.

3. Final Answer

$K = -\frac{1}{2} - \ln(2)$