$\iint (x+y) dx dy $ What is my mistake? Question
Solve the indefinite integral: $\iint (x+y) dx dy $
Attempt
When calculating this indefinite double integral, I would first start with x and then y such that my solution would be:

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*$\frac{1}2x^2 y + \frac{1}2y^2 x + c_1y + c_2$
But Wolfram Alpha's Solution is:

*

*$\frac{1}2x^2 y + \frac{1}2y^2 x + c_1x + c_2$
What am I doing wrong?
Thanks in advance!
 A: In one dimension, if the indefinite integral of $f(x)$ is $F(x)+c$, this means that when you differentiate $F(x)+c$ you get back to $f(x)$. $F(x)+c$ really denotes a family of functions, the antiderivatives of $f$, which differ from each other only by a constant; when you differentiate any of these antiderivatives, the $c$ disappears, so you always get $f$ as a result.
If we extend this to two dimensions, we find that the antiderivatives of $f(x,y)$ take the form $F(x,y)+c_1(x)+c_2(y)$, where $c_1$ and $c_2$ are arbitrary functions of one variable. This is because when we differentiate this with respect to $x$ and $y$, i.e. $\frac{\partial^2}{\partial x\partial y}(F(x,y)+c_1(x)+c_2(y))$, the functions $c_1$ and $c_2$ disappear in either the first or the second stage, so the result is independent of the functions $c_1$ and $c_2$. (I am ignoring complications that arise when the order of differentiation matters; this won't happen as long as the partial derivatives of $f$ are continuous.)
Your answer differs from Wolfram Alpha's answer by a function of the form $ax+by+c$; and because this function can be expressed as the sum of a function only of $x$ and a function only of $y$, they are equally valid antiderivatives of the function $f(x,y)=x+y$.
But a more complete answer would be $\frac12 x^2y+\frac12 y^2x + c_1(x)+c_2(y)$.
