[剑桥/9709/2007] 求常数 $k$ 的精确值，使得 $\int_{1}^{k} \frac{1}{2x - 1} \, dx = 1$。

1. Find the indefinite integral of the given function.

$\int \frac{1}{2x - 1} \, dx = \frac{1}{2}\ln(2x - 1)$

2. Evaluate the definite integral using the upper limit $k$ and lower limit $1$.

$\left[ \frac{1}{2}\ln(2x - 1) \right]_1^k = 1$
$\frac{1}{2}\ln(2k - 1) - \frac{1}{2}\ln(2(1) - 1) = 1$

3. Simplify the equation using the value of $\ln(1)$.

Since $\ln(1) = 0$, the equation simplifies to:
$\frac{1}{2}\ln(2k - 1) - 0 = 1$
$\frac{1}{2}\ln(2k - 1) = 1$

4. Solve the logarithmic equation to find the exact value of $k$.

$\ln(2k - 1) = 2$
$2k - 1 = e^2$
$2k = e^2 + 1$
$k = \frac{1}{2}(e^2 + 1)$