# Why there's a different answer for $\int x\ln{x^2}dx$ in wolfram alpha?

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.

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

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

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