Problem: $\int x\ln x^2 \,dx$
So what I did first was make $u = \ln x^2$ and $dv = x$
Then I solved by getting the derivative of $u$ and the anti derivative of $dv$ and I got $du = 1/x^2 $ and $v = x^2/2$ then I did the formula $$\int udv = uv - \int vdu$$ which then after plugging in the numbers and simplifying got me
$$ \frac{ x^2}{2}\ln x^2 - \frac{1}{2x} +C$$
Is this the right way to do the problem and answer?