Question: $\left[\log ({x^2+y^2}) + \left(\frac{2x^2}{x^2+y^2}\right)\right]dx$ + $\left(\frac{2xy}{x^2+y^2}\right)dy$
So I found $\left(\frac{\delta(m)}{\delta(y)}\right)$ and $\left(\frac{\delta(n)}{\delta(x)}\right)$, where m = $[log ({x^2+y^2}) + \left(\frac{2x^2}{x^2+y^2}\right)]$ and n = $\left(\frac{2xy}{x^2+y^2}\right)$, and I got $\left(\frac{2y^3-2x^2y}{(x^2+y^2)}\right)$ and $\left(\frac{2y^3-2x^2y}{(x^2+y^2)}\right)$ respectively, i.e $\left(\frac{\delta(m)}{\delta(y)}\right)$ = $\left(\frac{\delta(n)}{\delta(x)}\right)$, which means this is an exact differential equation. Now I have to find $\int(m)dx$ considering y terms as constants and $\int$(terms of n not containing x)dy = c. I am unable to proceed from here on. Any help will be appreciated. Please and thank you.
I assume you are trying to find $u(x,y)$ such that $$ du = \left[\log ({x^2+y^2}) + \left(\frac{2x^2}{x^2+y^2}\right)\right]dx +\left(\frac{2xy}{x^2+y^2}\right)dy. \tag{1} $$ If so, we can find it as follows: comparing $(1)$ with $$ du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy, \tag{2} $$ we have \begin{align} \frac{\partial u}{\partial y} = \left(\frac{2xy}{x^2+y^2}\right) &\Rightarrow u(x,y) = x\log(x^2+y^2) + f(x) \\ &\Rightarrow \frac{\partial u}{\partial x} = \log(x^2+y^2) + \frac{2x^2}{x^2+y^2} + f'(x). \tag{3} \end{align} Comparing $({3})$ with the coefficient of $dx$ in $(1)$, we conclude that $f'(x)=0$, so $f(x)=C$ and, finally, $$ u(x,y) = x\log(x^2+y^2) + C, \tag{4} $$ where $C$ is an arbitrary constant.