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

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.

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You want to find $f$ and $g$ such that $a=(fg)'$? Could you give an example? –  Pedro Tamaroff Jul 11 '12 at 23:49
@PeterTamaroff: Let $a(x)=\cos^3(x)-\sin^2(x)\cos(x)$. Then we can pick an $f(x) = \sin(x)$ and $g(x)=\cos^2(x)$. It's easy to come up with an example if you start with $f(x)$ and $g(x)$. But I find it tough to go the other way. –  Matt Groff Jul 12 '12 at 0:10
It can't be any easier than antidifferentiating $a$, since, if you can do it, you can antidifferentiate $a$ just by writing down $fg$. –  Gerry Myerson Jul 12 '12 at 0:51
It's as hard as integrating. If you can find $f,g$, then $fg$ is an indefinite integral. If you can find an indefinite integral $f$, then you can use $g = 1$. –  Yuval Filmus Jul 12 '12 at 1:42
I like to think of this as the same as applying integration by parts, since it comes from $d(uv) = d(u)v + u(dv)$. –  Thomas Belulovich Aug 13 '12 at 18:44

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.

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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

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

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How do you find $f(x)$ and $g(x)$, without knowing the solution to the integral? –  Matt Groff Aug 13 '12 at 19:58
The way integration by parts works is to transform $\int a {\rm d}x$ into $\int u {\rm d}v$ and then computing the r.h.s. of the equality above. –  ja72 Aug 13 '12 at 20:56