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

Question

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

Answer 1

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