integration of double integral with euler number include it. Hello i have a hard time with integration of double integral with euler number with power included in it. I have $$\int_0^1\int_0^2x^2ye^{xy}dxdy$$ so i am trying to use substitution for solving it. as follows i say $u = y$ $du = 1$ $v=ye^{xy}$ i am not sure for $dv$ here is it $e^{xy}$ or it is $\frac{e^{xy}}{x}$ ? Then i use the $Integration$ $by$ $parts$ $formula$ $$\int_a^budv=uv|_a^b-\int_a^bvdu$$ so from here i have $$\int_0^1x^2ye^{xy}|_0^2-\int_0^2x^2\frac{e^{xy}}{x}$$ a first thing i notice here i am left with 2 integrals one with respect to $dx$ and another with respect to $dydx$ that is confusing me in the first place so i am continue to integrate now i think i don't need $u$ $substitution$ to finish the integral with respect to $dy$ it must be : $$\int_0^1(2e^{2x}|_0^2-e^{2x}+e^0 )dx$$ From here i must do another $u$ $sub$ for x this time. So i am having many steps that i am not sure that are correct if someone can help me to understand whats happening here. Thank you in advance.

Let's do the integration in the following order:
$$\int_0^1\int_0^2x^2ye^{xy}dxdy=\int_0^2x^2\int_0^1ye^{xy}dydx.$$
The integral of the internal part (integrating by parts) is
$$\int_0^1 ye^{xy}dy=e^{xy}\frac{xy-1}{x^2}\big|_0^1=e^x\frac{x-1}{x^2}+\frac{1}{x^2}=\frac{xe^x-e^x+1}{x^2}.$$
When doing the outer integral  $x^2$ cancels out. That is, we have
$$\int_0^2xe^xdx-\int_0^2e^xdx+\int_0^21dx=e^x(x-2)+x\big |_0^2=4.$$
EDIT
We'll do the following indefinite integral by parts:
$$\int ye^{xy}dy.$$
Let $u=y$ and $v'=e^{xy}$. Then $v=\frac{e^{xy}}{x}$. So
$$\int ye^{xy}dy= y \frac{e^{xy}}x-\int \frac{e^{xy}}xdy=y \frac{e^{xy}}x-\frac{e^{xy}}{x^2}=e^{xy}\frac{xy-1}{x^2}.$$
