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A Quotient Rule Integration by Parts Formula

A Quotient Rule Integration by Parts Formula. Jennifer Switkes ([email protected]), California State Polytechnic Univer- sity, Pomona, CA 91768 In a recent calculus course, I introduced the technique of Integration by Parts as an integration rule corresponding to …

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calculus – Is there a rule of integration that corresponds

This quotient rule can also be deduced from the formula for integration by parts. The new formula is simply the formula for integration by parts in another shape. Therefore it has no new information, but its form allows to see what is needed for calculating the integral of the quotient of two functions.

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.Best answer · 16I guess you could arrange an analog to integration by parts, but making students learn it would be superfluous. $$ \int \frac{du}{v} = \frac{u}{v} + \int \frac{u}{v^2} dv.$$16It’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.14Is 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 \begin{equation}

\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}

\end{equation} 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 \begin{equation}

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}

\end{equation} 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.6For integrating a quotient of two functions, usually the rule for integration by parts is recommended:

\begin{equation}\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx,\end{equation}

\begin{equation}\int f'(x)g(x)dx=f(x)g(x)-\int f(x)g'(x)dx.\end{equation}

You have to choose $f$ and $g$ so that the integrand at the left side of one of the both formulas is the quotient of your given functions. But that needs some experience, and for the inexperienced or no-more skilled, this is somewhat unwieldy. In “A Quotient Rule Integration by Parts Formula” and in “Quotient-Rule-Integration-by-Parts” , the authors integrate the quotient rule of differentiation and get so a quotient rule integration by parts formula, where $F(x)=\int f(x) dx+c_{1}$, $c_{1}$ a constant:

\begin{equation}\int\frac{f'(x)}{g(x)}dx=\frac{f(x)}{g(x)}+\int f(x)\frac{g'(x)}{g(x)^{2}}dx,\end{equation}

\begin{equation}\int\frac{f(x)}{g'(x)}dx=\frac{F(x)}{g'(x)}+\int F(x)\frac{g”(x)}{g'(x)^{2}}dx.\end{equation}

Just as we get the quotient rule for differentiation from the product rule, we get this quotient rule for integration from the rule for integration by parts. Because these formulas still do not look like true quotient rules, I brought them to the following form, where again $F(x)=\int f(x) dx+c_{1}$, $c_{1}$ a constant: \begin{equation}\int\frac{f(x)}{g(x)}dx=\frac{F(x)}{g(x)}+\int F(x)\frac{g'(x)}{g(x)^{2}}dx,\end{equation} \begin{equation}\int_a^b\frac{f(x)}{g(x)}dx=\frac{F(x)}{g(x)}\bigg\vert_{a}^{b}+\int_a^b F(x)\frac{g'(x)}{g(x)^{2}}dx.\end{equation} This quotient rule can also be deduced from the formula for integration by parts. The new formula is simply the formula for integration by parts in another shape. Therefore it has no new information, but its form allows to see what is needed for calculating the integral of the quotient of two functions. I derived an anlog formula for the product rule of integration in “Are the real product rule and quotient rule for integration already known?” . Recently, this quotient rule of integration was also published in Will, J.: Produktregel, Quotientenregel, Reziprokenregel, Kettenregel und Umkehrregel für die Integration. May 2017 . The experienced will use the rule for integration of parts, but the others could find the new formula somewhat easier. It can be assumed that other quotient rules are possible. That means, the right side of the quotient rule can be written also in different forms.1

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