Prove $E(XY|X) = XE(Y|X)$
$f_{X,Y}(x,y)$ be the Joint density of $X$ and $Y$
Then by Jacobian Transform, joint density of $XY,X$ is $f_{XY,X}(z,x)=f_{X,Y}(x,\frac zx)\frac1{|x|}$
$$E[XY|X]=\int_{\mathbb R}zf_{XY|X}(z|x)dz={\int_{\mathbb R}zf_{X,Y}(x,z/x)|x^{-1}|dz\over f_X(x)}$$
Applying the substitution $z=xy$ with $y$ as the running variable
$$=f_X(x)^{-1}\int_\mathbb R x^2|x|^{-1}yf_{X,Y}(x,y)dy=|X|E[Y|X] \ne XE[Y|X]$$
Can someone point out the error? Thank you
By the way, the integrals you are calculating are things like $E[XY\mid \color{blue}{X = x}]$, which is a constant (since $x$ is a constant), rather than $E[XY\mid X]$, which is a random variable.
Anyway, the reason for the apparent discrepancy: take cases of whether $x < 0$ or $x > 0$ (where $x$ is fixed). If $x < 0$, your integral $\int_{-\infty}^{\infty}$ will become $\color{red}{\int_{\infty}^{-\infty}}$ after you do the substitution (bounds flip). This will then become $\int_{-\infty}^{\infty}=\int_{\Bbb{R}}$ if you introduce a minus sign, which will end up giving you a $-|x|$, which equals $x$ (as $x < 0$). If $x >0$, $|x|=x$ already.