5. \( \int^{\frac{1}{2}} \frac{1}{y(y-1)} d y \)

To solve the integral \( \int^{\frac{1}{2}} \frac{1}{y(y-1)} d y \), we can use partial fraction decomposition. We can express \( \frac{1}{y(y-1)} \) as: \[ \frac{1}{y(y-1)} = \frac{A}{y} + \frac{B}{y-1} \] Multiplying through by \( y(y-1) \) gives: \[ 1 = A(y-1) + By \] Setting \( y = 1 \) gives \( A(1-1) + B(1) = 1 \Rightarrow B = 1 \). Setting \( y = 0 \) gives \( A(0-1) + B(0) = 1 \Rightarrow -A = 1 \Rightarrow A = -1 \). Thus, we have: \[ \frac{1}{y(y-1)} = -\frac{1}{y} + \frac{1}{y-1} \] Now we can integrate: \[ \int \left(-\frac{1}{y} + \frac{1}{y-1}\right) dy = -\ln|y| + \ln|y-1| + C \] Evaluating the definite integral from \( 0 \) to \( \frac{1}{2} \): \[ \left[-\ln|y| + \ln|y-1|\right]^{\frac{1}{2}}_{0} \] At \( y = \frac{1}{2} \): \[ -\ln\left(\frac{1}{2}\right) + \ln\left(\frac{1}{2}-1\right) = -\ln\left(\frac{1}{2}\right) + \ln\left(-\frac{1}{2}\right) \] At \( y = 0 \): The natural logarithm diverges negatively, indicating an improper integral. Let's consider just the finite limit contributions: \[ \text{As } y \rightarrow 0, -\ln|y| \rightarrow +\infty \] Thus, the total integral diverges. Therefore, we conclude that: \[ \int^{\frac{1}{2}} \frac{1}{y(y-1)} d y = \infty \]