I'm attempting to solve $\int x\ln{x^2}dx$

Using integration by parts I was able to do the following steps.

$$2\int x \ln x dx=2\left(\frac{x^2}{2}\ln x -\int\frac{x^2}{2}dx\right)=x^2\ln x - \frac{x^3}{3}+c$$

But when I verified it with wolfram alpha, I'm getting a different answer.

i.e. $\frac{1}{2}x^2(\log x^2 -1)+c$

Can anyone please explain me why there's a difference? Thank you.

wolfram alpha link: https://www.wolframalpha.com/input/?i=integrate+xln%28x%5E2%29

  • 1
    $\begingroup$ yes. thanks for pointing that out. I've edited it. $\endgroup$
    – emil
    Dec 19, 2020 at 16:44

1 Answer 1


There is an error; it should be $$\int x\ln x^2 dx \overset{t=x^2}= \frac12\int \ln t dt= \frac12(t\ln t-t)+C$$

  • $\begingroup$ Can't I apply logarithm rule to it to take the power as a linear factor? $\endgroup$
    – emil
    Dec 19, 2020 at 14:58
  • $\begingroup$ $\ln{x^r}=r\ln{x}$? $\endgroup$
    – emil
    Dec 19, 2020 at 14:59
  • $\begingroup$ @emil - you could. Then, it should be $$2\int x\ln x dx = \int \ln x d(x^2) = x^2\ln x -\int xdx =x^2\ln x-\frac12 x^2$$ $\endgroup$
    – Quanto
    Dec 19, 2020 at 15:01
  • 1
    $\begingroup$ oh yes! I found my mistake. Thank you!! $\endgroup$
    – emil
    Dec 19, 2020 at 15:03
  • 2
    $\begingroup$ @emil: You should mark this answer as accepted so that it doesn't queue in the Unanswered. $\endgroup$ Dec 19, 2020 at 15:12

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