Product rule says that $(uv)' = u'v + uv'$, so $\int (uv)' = \int (u'v + uv')$ implies $uv = \int u'v + \int uv'$ and this implies

$$\int uv' ~dx = uv - \int u'v ~dx$$

This is integration by parts. I am wondering if this also works with quotient rule:

$(u/v)' = \frac{vu' - uv'}{v^2}$ so $\int(u/v)' = \int\frac{vu' - uv'}{v^2}$ implies $u/v = \int\frac{vu'}{v^2} - \int\frac{uv'}{v^2}$ and also

$$\int\frac{u'}{v} ~ dx = u/v + \int\frac{uv'}{v^2} ~ dx$$

I am not sure if this relationship would have any uses but would it be a valid method?

  • 5
    $\begingroup$ This looks equivalent to integration by parts, just that you replaced $v \rightarrow u$ and $u \rightarrow \frac{1}{v}$. $\endgroup$
    – Element118
    Jan 16 '16 at 16:28
  • 3
    $\begingroup$ The quotient rule is often derived by using the product rule so as @Element118 notes, they are basically the same, but definitely a +1 for wondering this! $\endgroup$ Jan 16 '16 at 16:30
  • 1
    $\begingroup$ You may care to take a look at this paper. $\endgroup$
    – omegadot
    Jan 17 '16 at 9:03

Hint : Both are the same for all $v \neq 0$

Replace $v$ by $\frac{1}{v}$ .


For 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,
Will, J.: Product rule, quotient rule, reciprocal rule, chain rule and inverse rule for 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.