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:


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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – user958619
    Jan 8, 2022 at 7:40
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    $\begingroup$ 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. $\endgroup$ Jan 8, 2022 at 7:56
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    $\begingroup$ 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. $\endgroup$ Jan 8, 2022 at 8:04
  • $\begingroup$ Thanks @DanielWainfleet for your edits! It looks much better. $\endgroup$ Jan 9, 2022 at 9:16

1 Answer 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 $\frac{15}{8k}$


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