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