# Different methods of integration yield different results?

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!

• The second part of the integration should be $3ln\left(8\right)+3ln\left(x\right)$ which is correct while your answer which is $ln\left(8\right)+3ln\left(x\right)$ is wrong.
– user958619
Jan 8, 2022 at 7:40
• My edit was for appearance. It is better to break a very long sequence between dollars into smaller parts that each fit on a single line. And the modern style is that if a line ends with a relational symbol such as $=$ then the symbol is repeated at the start of the next line. I see you found your computation error OK. Jan 8, 2022 at 7:56
• You can use \ln and \log in LaTex, e.g. \ln x yields $\ln x$ which means you don't need so many brackets.... The customary symbol here for the real numbers is usually $\Bbb R$, LaTex \Bbb R. Jan 8, 2022 at 8:04
• Thanks @DanielWainfleet for your edits! It looks much better. Jan 9, 2022 at 9:16

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