Different methods of integration yield different results?
Solution 1:
It turns out I have made a mistake.
$$(\frac{\ln(8)}{2k} +\frac{3((\frac{3}{2}-\ln(2))^2-(\frac{1}{2}-\ln(2))^2)}{2k} ) $$
This step is incorrect as $\ln(\frac{e}{2})=1-\ln(2)$, rather than $\frac{1}{2}-\ln(2)$. As a result, the definite integral is not evaluated correctly. Using the correct expression:
$$(\frac{\ln(8)}{2k} +\frac{3((\frac{3}{2}-\ln(2))^2-(1-\ln(2))^2)}{2k} ) $$
The terms for $\ln(2)$ cancel out and, indeed, we are left with only $\frac{15}{8k}$