Question

what is the integral of (4-x-y)dx.

Answer 1

\[4x - {x^2 \over 2} - yx + C\]

Answer 2

I would write \[4x-\frac{x^2}{2}-yx+C(y)\]

where \(C(y)\) is just a function of \(y\)

since \(\displaystyle \frac{\partial}{\partial x}\left[4x-\frac{x^2}{2}-yx+C(y)\right]=4-x-y\)