Are there functions satisfying $\int f(x)g(x)dx=\int f(x)dx \times \int g(x)dx$
I came up with one which is
$f(x)=e^{x \sec ^2 \alpha}$ and $g(x)=e^{x \csc^2 \alpha}$
$$\int f(x)g(x)dx=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha +\csc ^2 \alpha}$$
Also
$$\int e^{x\sec^2 \alpha}dx \times \int e^{x \csc^2 \alpha} dx=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha \csc ^2 \alpha}=\frac{e^{x(\sec^2 \alpha+\csc^2 \alpha)}}{\sec^2 \alpha +\csc ^2 \alpha}$$
Let $F = \int f, G = \int g $.
I'll try to get $f$ in terms of $g$.
Differentiating $\int fg =\int f \int g $, $fg =g\int f+f\int g =g\int f+fG $ or, in terms of $F$, $F'(g-G) =gF$ so $\dfrac{F'}{F} =\dfrac{g}{g-G} $ or $(\ln(F))' =\dfrac{g}{g-G} $ or $\ln F =\int(\dfrac{g}{g-G}) $ so, $ F =e^{\int(\dfrac{g}{g-G})} $ and, differentiating, $f =\dfrac{g}{g-G}e^{\int(\dfrac{g}{g-G})} $.
Example:
Let $g = x^n$ so $G = \dfrac{x^{n+1}}{n+1} $. $\dfrac{g}{g-G} =\dfrac{x^n}{x^n-\dfrac{x^{n+1}}{n+1}} =\dfrac{1}{1-\dfrac{x}{n+1}} =\dfrac{n+1}{n+1-x} $ so $\int\dfrac{g}{g-G} =\int\dfrac{n+1}{n+1-x} = -(n + 1) \log(n - x + 1) $ and $f =\dfrac{n+1}{n+1-x}e^{-(n + 1) \log(n - x + 1) } =\dfrac{n+1}{n+1-x}\dfrac1{(n-x+1)^{n+1}} =\dfrac{n+1}{(n-x+1)^{n+2}} $.