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I've seen occasions such as the xkcd comic

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Or the classic - simplifying $16/64$ by cancelling out the six's to get $1/4$.

I was grading some intro-to-calc papers recently, and one student did something like the following:

$\int x\sqrt{1 + x^2} = \frac{x^2}{2} \sqrt{ 1 + \frac{x^3}{3} } + C$

How hard would it be to construct an integral of a function such that doing this 'naive' sort of integration (I suppose we are distributing the integral operator, in particular) will actually return the correct answer? I played around briefly, and it seemed like an interesting challenge.

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    I have the distinct impression of having dealt with the problem of when the integral of a product is the product of the integrals (or when the derivative of the product is the product of the derivatives) sometime before, but I can't seem to find it either here or in sci.math...2012-02-23

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You basically want functions $f$ and $g$ such that (fg)' = f'g'. If $f=0$ then this is trivially true, but not what you want. If g=g', then we have f'g + fg' = f'g' which yields (f'+f)g = f'g, hence $g=0$ or $f=0$.

Assuming g\neq g' and $f\neq 0$, this leads to f'g + fg' = f'g', or \frac{f'}{f} = \frac{g'}{g'-g} so you want to find an $f$ such that \ln|f| = \int\frac{g'}{g'-g}\,dx.

Pick any $g$ for which this can be done easily. For instance, say $g=x$. Then g'=1, so we have $ \ln|f| = \int\frac{dx}{1-x} = -\ln|x-1| + C,$ so $f(x) = \frac{A}{x-1}$, $g(x)=x$ will do it. Indeed, \left(\frac{Ax}{x-1}\right)' = \frac{(x-1)A - Ax}{(x-1)^2} = -\frac{A}{(x-1)^2}, and \left(\frac{A}{x-1}\right)'\left(x\right)' = -\frac{A}{(x-1)^2}.

So just pick your favorite $g$ for which \int\frac{g'}{g'-g}\,dx is "doable" and you get an example.

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    Of course, the example above is "uninteresting" since $\int f'g'\,dx = \int f'$, so it is hard to recognize the function as a product. But you can use other $g$s, like $g(x) = \sqrt{x}$, or even $g(x) = \sin x$...2012-02-23