There is a problem :
$\int x \ln(x^2)\,dx = \int 2x \ln x\,dx$
How can I simplify the integral like in the right?
I tried to use substitution
$x^2=u$,
$2x=\frac{du}{dx}$
$\int x \ln(x^2)\,dx= \int\frac{u}{2} \ln u\, du$,
so its not the same?
Can someone point out what is wrong with my attempt?
Recall that
$$\ln x^2=2 \ln x$$
note that by substitution $x^2=u \implies 2xdx=du\implies dx=\frac1{2\sqrt u}du$ we obtain
$$\int x \ln(x^2)dx= \int \sqrt u\ln u \frac1{2\sqrt u}du=\frac12\int \ln u \,du$$