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I have a partial derivative $f_x(x,y)=-\sin(y)+\frac1{(1-xy)}$ and $f(0,y)= 2\sin(y)+y^3$

I try to take an integral like this:

$$\int -\sin(y)+\frac1{1-xy}dx=-\frac{1}{y}\ln \left|1-yx\right|-\sin \left(y\right)x$$

After I plug in the $(0,y)$, the answer is $0$.

Whats wrong with my formula? And how to compute the original $f(x,y)$ with partial derivatives?

Gary
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Joyce
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    Why does your antiderivative lack a function of integration, $C(y)$ (similar to a constant of integration, but here, differentiation by $x$ sends any function of $y$ (only) to zero, so we get an arbitrary function of $y$, not just an arbitrary constant)? See https://math.stackexchange.com/a/754753/123905 for an example. – Eric Towers Oct 15 '21 at 07:21
  • Thank you for the notice! It seems that I miss the constant C. – Joyce Oct 15 '21 at 07:40

1 Answers1

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You have$$f(x,y)-f(0,y)=\int_0^xf_x(t,x)\,\mathrm dt;$$in other words,$$f(x,y)=2\sin(y)+y^3+\int_0^x-\sin(y)+\frac1{1-ty}\,\mathrm dt.$$

José Carlos Santos
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