Different methods of integration yield different results

calculusdefinite integralsintegrationsubstitution

The problem I must solve for a high-school math paper is the value of the definite integral:
$$\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8x^3)}{kx}dx, k\in \mathbb{R}^+$$

Not looking too hard at first glance, I solved it as such:
\begin{align} \int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8x^3)}{kx}dx &=\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8)+3\ln(x)}{kx}dx\\ &=\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8)}{kx}dx+\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{3\ln(x)}{kx}dx\\ &=(\frac{\ln(8)\ln(x)}{k} +\frac{3\ln^2(x))}{2k} ) \Big|_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\\ &=(\frac{\ln(8)}{2k} +\frac{3((\frac{3}{2}-\ln(2))^2-(\frac{1}{2}-\ln(2))^2)}{2k} ) \\ &=\frac{6-3\ln(2)}{2k} \end{align}

However, the "technically correct" way of solving the integral is by direct u-substitution.

This is done by setting $$u=\ln(8x^3)$$ and then $$du=\frac{3}{x}dx$$. Then $$\frac{1}{kx}dx = \frac{1}{3k}du$$ and hence:

$$\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8x^3)}{kx}dx=\frac{1}{3k}\int_{3}^{\frac{9}{2}}udu=\frac{15}{8k}$$

These results are obviously distinct from each other, as $$4\times(6-3\ln(2)) \neq 15$$. It confuses me why my method, although different from the expected method, obtains the incorrect answer. I also can't seem to find any mistakes in my own working out.

Any help will be greatly appreciated. Thanks!

$$(\frac{\ln(8)}{2k} +\frac{3((\frac{3}{2}-\ln(2))^2-(\frac{1}{2}-\ln(2))^2)}{2k} )$$
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 $$\frac{15}{8k}$$