I am having some trouble with the following integral:

$$\int \frac{\ln(15x^5)}{x}$$

I separated the top into:

$$\int \frac{5\ln(x)+ \ln(15)}{x}$$

But, then I don't know where to go from there. I tried $u$ substitution by letting $u=\ln(x)$.

\begin{align} \int 5u &+ \ln (15) \, du \\ \frac{5u^2}{2} &+ \ln(15)u \\ \frac{5(\ln(x))^2}{2}&+\ln(15)\ln(x) \end{align}

Where am I going wrong? Can someone please provide some hints?

Thanks a bunch!

  • 1
    $\begingroup$ $\ln xy=\ln x+\ln y\neq \ln x\cdot \ln y$ $\endgroup$
    – L. F.
    Aug 15, 2013 at 22:47
  • $\begingroup$ Whoops! I transferred that incorrectly from my paper to the question. $\endgroup$
    – Jeel Shah
    Aug 15, 2013 at 22:48
  • $\begingroup$ There is nothing wrong - just don't forget the arbitrary constant as well! $\endgroup$
    – L. F.
    Aug 15, 2013 at 22:51
  • 1
    $\begingroup$ You could also find this integral by substituting $u=\ln (15x^5)$, but I like your method better. $\endgroup$
    – user84413
    Aug 15, 2013 at 22:57
  • 1
    $\begingroup$ @gekkostate That's the same, modulo the additive constant. $(5\cdot\ln(x)+\ln(15))^2 = 25(\ln x)^2 + 10\ln x \ln 15 + (\ln 15)^2$. $\endgroup$ Aug 15, 2013 at 22:58

1 Answer 1


$$\begin{align}\int \frac{\ln(15x^5)}{x}\,dx & = \int \frac{5\ln(x)+ \ln(15)}{x}\,dx \\ \\ & = 5\int \frac{\ln x}{x} dx + \int \ln(15)x^{-1} \,dx\end{align}$$

You integrated the first portion correctly: $u = \ln x \implies du = \frac 1x\,dx$. And it so happens that you got the second integral correct too. (Indeed, there is no real need to separate the summed integrand into the sum of two integrals.)

Recall that $\ln(15)$ is nothing more than a constant.

There is some simplification you can do, if desired (using properties of the logarithm function). But your integration is just fine.

We could, alternatively, start from the beginning and set $$u = \ln(15x^5) \implies du = \dfrac{5\cdot 15x^4}{15x^5} \,dx = \frac 5x\,dx$$

$$\int \frac{\ln(15x^5)}{x}\,dx = \frac 15\int u\,du$$

and then go from there.

  • $\begingroup$ I haven't learned integration by parts. All I know is $u$ substitution and sorry about the mistake in the post. I transferred incorrect from paper to question. $\endgroup$
    – Jeel Shah
    Aug 15, 2013 at 22:50
  • $\begingroup$ No need here, sorry about that. $\endgroup$
    – amWhy
    Aug 15, 2013 at 22:50
  • $\begingroup$ Glad to know that the integration was correct :). Thanks for the help! $\endgroup$
    – Jeel Shah
    Aug 15, 2013 at 23:04
  • $\begingroup$ @amWhy: You have very good patience too! +1 $\endgroup$
    – Amzoti
    Aug 16, 2013 at 12:54
  • $\begingroup$ @amWhy: I agree with Amzoti, too +) $\endgroup$
    – Mikasa
    Aug 16, 2013 at 15:22

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