Is there a rule of integration that corresponds to the quotient rule?

When teaching the integration method of u-substitution, I like to emphasize its connection with the chain rule of integration.
Likewise, the intimate connection between the product rule of derivatives and the method of integration by parts comes up in discussion.

Is there an analogous rule of integration for the quotient rule?

Of course, if you spot an integral of the form $\int \left (\frac{f(x)}{g(x)} \right )' = \int \frac{g(x) \cdot f(x)' - f(x) \cdot g(x)'}{\left [ g(x)\right ]^2 }$,
then the antiderivative is obvious. But is there another form/manipulation/"trick"?

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Note that $\frac fg = f\cdot {\frac1g}$, so there is in fact no new rule for the quotient. The only rule is for product. – Ilya Sep 29 '11 at 12:28
Yeah, nothing really that new. FWIW, I always had derived the quotient rule from both the product rule and chain rule, so I'd say those two are more fundamental, and you've covered those already. In any event... "product rule of integration"? – J. M. Sep 29 '11 at 12:37
@J. M. - it was a test to see who had their coffee this morning. That reminds me... :) – The Chaz 2.0 Sep 29 '11 at 12:39
@TheChaz: I believe our location is different. On the North Sea beach it's now afternoon and very hot, so coffee doesn't help much (( – Ilya Sep 29 '11 at 12:46
@Gortaur: Next time I'm up there, I'll buy a round of iced coffees! But seriously, would you care to (slightly) elaborate on your comment and put it as an answer? – The Chaz 2.0 Sep 29 '11 at 12:48

As for me, I cannot see an advantage in introduction of such a rule since for any two functions $f,g$ it clearly holds that $$\frac fg = f\cdot\frac1g$$ so the 'quotient rule' for derivatives is a product rule in disguise, and the same will also hold for the integration by parts. Indeed, when you are looking for the proper function to put under the differential sign integrating by parts, in case you have a bit of experince with such a procedure, you also will think about the 'quotients'.

As an example: $$\int\frac{\sin\frac1x}{x^2}\,dx$$ Of course you can present it as $\frac{f(x)}{x^2}$ and apply the new integration by parts based on the quotient rule, but I almost sure that a lot of the readers will rather think of the fact that $\frac1{x^2}\,dx = -d\frac1x$, by this seeing a product in the integrand rather than a quotient.

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I guess you could arrange an analog to integration by parts, but making students learn it would be superfluous.

$$\int \frac{du}{v} = \frac{u}{v} + \int \frac{u}{v^2} dv.$$

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I kind of like it, no minus signs. Also has nice natural form when one uses integration by parts (sorry, the Ault Rule) for estimation. – André Nicolas Sep 29 '11 at 13:54
Cool, I like it as well for the same reason of the absence of minuses. – Ilya Sep 29 '11 at 13:55

It's worth emphasizing that a "quotient rule" does play a role in Hermite's algorithm for integrating rational functions. It works as follows. By squarefree decomposing the denominator and partial fraction expanding, we reduce to integrating $\rm\:A/D^k\in \mathbb Q(x)\:,\:$ where $\rm\:\deg\:A < \deg\:D^k,\:$ and where $\rm\:D\:$ is squarefree, so $\rm\:\gcd(D,D') = 1\:.\:$ Thus by Bezout (extended Euclidean algorithm) there are $\rm\:B,C\in \mathbb Q[x]\:$ such that $\rm\ B\ D' + C\ D\ =\ A/(1-k)\:.\:$ Then a little algebra shows that

$$\rm\int \frac{A}{D^k}\ =\ \frac{B}{D^{k-1}}\ +\ \int \frac{(1-k)\ C - B'}{D^{k-1}}$$

Iterating the above rule we eventually reduce to the case $\rm\:k=1\:,\:$ i.e. squarefree denominator $\rm\:D\:.\:$ Thus using the above "quotient rule" and nothing deeper than Euclid's algorithm for polynomials (without requiring any factorization) one can mechanically compute the "rational part" of the integral of a rational function, i.e. the part of the integral not involving logarithms. This Hermite reduction rule is the basis of an algorithm due to Hermite (1872). It plays a fundamental role in the transcendental case of some integration algorithms.

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In standard calculus courses,special cases of the Hermite procedure is usually called "the rule for integration by rational functions". I don't think it's really a "rule" in the usual sense we get in these courses,in that it's not an algorithm derived from using the Fundamental Theorum of Calculus to "invert" a derivative rule. It's just a handy application of Galois theory. – Mathemagician1234 Sep 29 '11 at 19:24
@Mat There is no Galois theory involved. It's not clear to me what restricted denotation of "rule" that you intend, nor why you think it plays a role here. I dont recall ever hearing of Hermite reduction being applied in any "standard calculus course". That would be quite some course. Do you have a link? – Bill Dubuque Sep 29 '11 at 19:44
I'll look,Bill. – Mathemagician1234 Sep 29 '11 at 22:22

Is there an analogous rule of integration for the quotient rule?

For a rational function $P(x)/Q(x)$, the answer is affirmative, and it's called the Ostrogradsky-Hermite method. I adapt this answer of mine.

To integrate a rational function $P(x)/Q(x)$ without decomposing it into partial fractions and without finding the roots of the denominator, we can use the Ostrogradski-Hermite method. You can find a description of this method in section 2.1 of Gradshteyn and Ryzhik's Table of Integrals, Series, and Products, where the identity $(2)$ bellow is given. The formula $(1)$ appears also on the Ostrogradsky's Wikipedia page.

Assume that $\deg P(x)<\deg$ $Q(x)$. There exist polynomials $P_{1}(x)$, $P_{2}(x)$, $Q_{1}(x)$ and $Q_{2}(x)$, with $Q_{1}(x)=\gcd \left\{ Q(x), Q^{\prime }(x)\right\}$ and $Q_{2}(x)=Q(x)/Q_{1}(x)$, $\deg P_{1}(x)<\deg Q_{1}(x)$, $\deg P_{2}(x)<\deg Q_{2}(x)$, such that

$$\int \frac{P(x)}{Q(x)}dx=\frac{P_{1}(x)}{Q_{1}(x)}+\int \frac{P_{2}(x)}{ Q_{2}(x)}dx.\tag{1}$$

Indeed, by differentiation and multiplication by $P(x)$, we have

\begin{eqnarray*} P(x) &=&\frac{P_{1}^{\prime }(x)Q_{1}(x)-P_{1}(x)Q_{1}^{\prime }(x)}{\left\{ Q_{1}(x)\right\} ^{2}}Q(x)+\frac{P_{2}(x)}{Q_{2}(x)}Q(x) \\ &=&P_{1}^{\prime }(x)\frac{Q(x)}{Q_{1}(x)}-P_{1}(x)\frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}\frac{Q(x)}{Q_{1}(x)}+P_{2}(x)\frac{Q(x)}{Q_{2}(x)} \\ &=&P_{1}^{\prime }(x)Q_{2}(x)-P_{1}(x)\left\{ \frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}Q_{2}(x)\right\} +P_{2}(x)Q_{1}(x) \end{eqnarray*}

or

$$P(x)=P_{1}^{\prime }(x)Q_{2}(x)-P_{1}(x)\left\{ T(x)-Q_{2}^{\prime }(x)\right\}+P_{2}(x)Q_{1}(x),\tag{2}$$

with $T(x)=Q^{\prime }(x)/Q_{1}(x)$, because from

\begin{equation*} Q^{\prime }(x)=\left\{ Q_{1}(x)Q_{2}(x)\right\} ^{\prime }=Q_{1}^{\prime }(x)Q_{2}(x)+Q_{1}(x)Q_{2}^{\prime }(x)=T(x)Q_{1}(x) \end{equation*}

we obtain

\begin{equation*} \frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}Q_{2}(x)+Q_{2}^{\prime }(x)=T(x). \end{equation*}

To find the coefficients of the polynomials $P_{1}(x)$ and $P_{2}(x)$ we equate the coefficients of like powers of $x$ and/or make $x=x_1,x_2,\dots$ until we obtain a system of linear equations in those coefficients.

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