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

AnalysisIntegrationDefinite IntegralsSubstitutionCalculus
2025/5/11

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)=2x(x1)2f(x) = -\frac{2-x}{(x-1)^2} can be written as f(x)=1(x1)2+1x1f(x) = -\frac{1}{(x-1)^2} + \frac{1}{x-1} and then to calculate K=10f(x)dxK = \int_{-1}^{0} f(x) dx.

2. Solution Steps

First, let's show that f(x)=2x(x1)2=1(x1)2+1x1f(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:
1(x1)2+1x1=1(x1)2+x1(x1)2=1+x1(x1)2=x2(x1)2=2x(x1)2-\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)f(x).
Next, we want to find K=10f(x)dx=10(1(x1)2+1x1)dxK = \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=101(x1)2dx+101x1dxK = \int_{-1}^{0} -\frac{1}{(x-1)^2} dx + \int_{-1}^{0} \frac{1}{x-1} dx.
For the first integral, let u=x1u = x-1. Then du=dxdu = dx. When x=1x = -1, u=2u = -2. When x=0x = 0, u=1u = -1.
101(x1)2dx=211u2du=21u2du=[u1]21=[1u]21=1112=1+12=12\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=x1u = x-1. Then du=dxdu = dx. When x=1x = -1, u=2u = -2. When x=0x = 0, u=1u = -1.
101x1dx=211udu=[lnu]21=ln1ln2=ln(1)ln(2)=0ln(2)=ln(2)\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=12ln(2)K = -\frac{1}{2} - \ln(2).

3. Final Answer

K=12ln(2)K = -\frac{1}{2} - \ln(2)

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