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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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    I like to think of this as the same as applying integration by parts, since it comes from $d(uv) = d(u)v + u(dv)$.2012-08-13

2 Answers 2

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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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    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.2012-08-13