Is symbolically solving $a(x)=f(x)g'(x)+f'(x)g(x)$, given $a$, ever easier than integrating $a$?

Question

If we are given a function of $x$, $a(x)$, how hard is it to find an $f(x)$ and $g(x)$ such that $$a(x)=f(x)g'(x)+f'(x)g(x)$$ For comparison, I'd like to know when this is easier than symbolically or numerically integrating $a(x)$.

I'd like to know, if possible, what general conditions allow us to efficiently find $f(x)$ and $g(x)$. I'm hoping this isn't too general a question. Additionally, I'd like to know the methods that allow us to do so.

Answer 1

Suppose $a(x)=x^2\cos x + 2x\sin x$.

Recognizing that as $f'(x)g(x)+ f(x)g'(x)$, with $f(x)= \sin x$ and $g(x)= x^2$, seems like the quickest way of finding the antiderivative of the whole expression.

Answer 2

Isn't this just integration by parts. Where

$$ \int a(x)\,{\rm d}x = f(x)g(x)= \int \frac{{\rm d}f(x) g(x)}{{\rm d}x}\,{\rm d}x = \int f(x)g'(x) \,{\rm d}x +\int f'(x)g(x) \,{\rm d}x $$ $$ = \int f(x) \,{\rm d}g +\int g(x) \,{\rm d}f $$

or

$$ \int u \;{\rm d}v = u\, v - \int v \;{\rm d}u $$