Are there functions satisfying $\int f(x)g(x)dx=\int f(x)dx \times \int g(x)dx$ 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}}
$.
