Question:

<p><strong>Step 1:</strong> Decompose using partial fractions.</p><p>$$\frac{1}{z^2-1} = \frac{1}{(z-1)(z+1)} = \frac{A}{z-1} + \frac{B}{z+1}$$</p><p>Multiplying by (z-1)(z+1): $$1 = A(z+1) + B(z-1)$$</p><p>Setting z=1: 1 = 2A → A = 1/2</p><p>Setting z=-1: 1 = -2B → B = -1/2</p><p><strong>Step 2:</strong> Rewrite the integral.</p><p>$$\int_2^3 \frac{dz}{z^2-1} = \int_2^3 \left(\frac{1/2}{z-1} - \frac{1/2}{z+1}\right)dz$$</p><p><strong>Step 3:</strong> Integrate.</p><p>$$= \frac{1}{2}\left[\ln|z-1| - \ln|z+1|\right]_2^3$$</p><p>$$= \frac{1}{2}\left[\ln\left|\frac{z-1}{z+1}\right|\right]_2^3$$</p><p><strong>Step 4:</strong> Apply limits.</p><p>$$= \frac{1}{2}\left[\ln\left(\frac{2}{4}\right) - \ln\left(\frac{1}{3}\right)\right]$$</p><p>$$= \frac{1}{2}\left[\ln\left(\frac{1}{2}\right) - \ln\left(\frac{1}{3}\right)\right]$$</p><p>$$= \frac{1}{2}\ln\left(\frac{3}{2}\right)$$</p><p>∴ Answer: C</p>