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I know the chain rule for derivatives. The way as I apply it, is to get rid of specific 'bits' of a complex equation in stages, i.e I will derive the $5$th root first in the equation $(2x+3)^5$ and continue with the rest.

I wonder if there is something similar with integration. I tried to integrate that way $(2x+3)^5$ but it doesn't seem to work. Well, it works in the first stage, i.e it's fine to raise in the power of $6$ and divide with $6$ to get rid of the power $5$, but afterwards, if we would apply the chain rule, we should multiply by the integral of $2x+3$!, But it doesn't work like that, we just need to multiply by $1/2$ and that's it.

So my question is, is there chain rule for integrals? I want to be able to calculate integrals of complex equations as easy as I do with chain rule for derivatives.

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    $\begingroup$ There is no direct equivalent, but the technique of integration by substitution is based on the chain rule. $\endgroup$ – StackTD Feb 1 '16 at 14:36
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    $\begingroup$ In fact there is not even a product rule for integration (which might seem easier to obtain than a chain rule). Even if you know primitives $F,G$ of respectively $f,g$, it is not guaranteed that you can find a primitive of their product $fg$. $\endgroup$ – Marc van Leeuwen Feb 4 '16 at 10:44

10 Answers 10

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The "chain rule" for integration is the integration by substitution.

$$\int_a^b f(\varphi(t)) \varphi'(t)\text{ d} t = \int_{\varphi(a)}^{\varphi(b)} f(x) \text{ d} x $$


So in your case we have $f(x) = x^5$ and $\varphi(t) = 2t+3$:

$$ \int (2t + 3)^5 \text{ d}t = \int {1 \over 2}\left((2t + 3)^5\cdot2\right) \text{ d}t = {1\over 2}\int x^5 \text{ d}x = {1\over 12} x^6 + C= {1\over 12} (2t+3)^6 + C$$

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    $\begingroup$ I don't agree to call this a chain rule for integration, which would be about the integrand $f(\phi(t))$, not $f(\phi(t))\phi'(t)$. What you have here is the chain rule for derivation taken backwards, nothing new. Check the answer by @GEdgar. $\endgroup$ – Yves Daoust Feb 3 '16 at 11:28
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    $\begingroup$ @YvesDaoust Guess why I put it in quotes? Integration with substitution is a way to deal with conposite functions. It's not a "rule" in that way it's always valid to get a solution as the chain rule for differentiation does. It's merely a "purpose equivalent". $\endgroup$ – adjan Feb 3 '16 at 12:07
  • $\begingroup$ Why is it $f(\phi(t))\phi'(t)$ not $f'(\phi(t))\phi'(t)$? $\endgroup$ – wkpk11235 Oct 26 '17 at 1:32
  • $\begingroup$ @wkpk11235 I know it's probably too late to comment this, but it is because we are only considering a case that reduces to $\int f$. We do not require that the integral gives us the function $f$, applied to endpoints. $\endgroup$ – Deepak M S Nov 16 '20 at 21:05
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If we know the integral of each of two functions, it does not follow that we can compute the integral of their composite from that information.

Example
$$ \int e^{-x}\;dx = -e^{-x} +C\\ \int x^2\;dx = \frac{x^3}{3} +C\\ $$ but $$ \int e^{-x^2}\;dx = \frac{\sqrt{\pi}}{2}\;\mathrm{erf}(x) + C $$ is not an elementary function.

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    $\begingroup$ This is the correct answer to the question. (Integration by substitution is not a chain rule, though it is certainly an important related technique.) But FTR, just because the error function is not elementary doesn't mean we can't compute it. $\endgroup$ – leftaroundabout Feb 2 '16 at 0:00
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    $\begingroup$ @leftaroundabout When did GEdgar say we can't compute it? $\endgroup$ – Simply Beautiful Art Feb 2 '16 at 22:21
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I'm guessing you're asking how to do the integral

$$\int (2x+3)^5 \, dx$$

I would use substitution:

$$u=2x+3 \\ du=2 \, dx$$

So your new integral is

$$\int \frac{u^5}2 \, du = \frac{u^6}{12} +C$$

Then you replace $u$ with the original $2x+3$ to get

$$\int \frac{u^5}2 \, du = \frac{u^6}{12} +C = \frac{(2x+3)^6}{12} +C$$

If you want to see how this relates to the chain rule, take the derivative of your answer, and it should get you the function "inside" the original integral.

$$F(x)=\frac{(2x+3)^6}{12} = f(g(x))$$ $$f(x)=\frac{x^6}{12} \, \, \, g(x)=2x+3 \\ f'(x)=\frac{x^5}2 \, \, \, g'(x)=2 \\$$

Using the chain rule we get

$$F'(x) = f'(g(x))g'(x) = f'(2x+3)g'(x) = \frac{(2x+3)^5}2 (2) = (2x+3)^5$$

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  • $\begingroup$ yeah but I am supposed to use some kind of substitution to apply the chain rule, but I don't feel the need to specify substitutes. I just solve it by 'negating' each of the 'bits' of the function, ie. first I go for the power if any, then I go for the rest bit, etc. I read in a stupid website that integration by substitution is ONLY to solve the integral of the product of a function with its derivative, is this true? Or you can solve ANY complex equation with that? do you have a good resource? $\endgroup$ – ergon Feb 1 '16 at 15:35
  • $\begingroup$ @addy2012 gave the formal definition for Integration by Substitution for a single variable, which is what I used in my answer. Integration by Substitution is the counterpart to the chain rule of differentiation. It is similar to how the Fundamental Theorem of Calculus connects Integral Calculus with Differential Calculus. You can take the derivative of the answer and you will get the $f(x)$ inside the original integral by applying the chain rule, which I will edit my answer to show. $\endgroup$ – Brendan Feb 1 '16 at 15:51
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    $\begingroup$ There is also integration by parts, which is almost like making two substitutions. It's a way of breaking down an integral into something you will be able to work with. u-substitution and Integration by Parts are probably some of the most useful tools you will use in Calculus I and II (assuming the common 3 semester separation). You can't solve ANY integral with just substitution, but it's a good thing to try first if you run into an integral that you don't immediately see a way to evaluate. $\endgroup$ – Brendan Feb 1 '16 at 15:57
  • $\begingroup$ If anyone can help me format my answer better I would really appreciate it, as I'm still learning the formatting (lining up the equals signs for $u$ and $du$ in the beginning as well as making the $f(x)$ $g(x)$ and their derivatives line up nicely). $\endgroup$ – Brendan Feb 1 '16 at 16:54
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    $\begingroup$ @ergon That website is indeed "stupid" (or at least unhelpful) if it really says that substitution is only to solve the integral of the product of a function with its derivative. It really is just running the chain rule in reverse. What makes this difficult is that you have to figure out which part of the integrand is $f'(g(x))$ and which is $g'(x)$. You can't just "chip away" one exponent/factor/term at a time as you can when differentiating. Sometimes the way is just to make what appears to be a likely guess based on similar integrals and see if it works. $\endgroup$ – David K Feb 1 '16 at 19:49
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No.

Sorry for turning up late here, but I think the other (excellent) answers miss a key point. There is no direct, all-powerful equivalent of the differential chain rule in integration. The existence of the chain rule for differentiation is essentially what makes differentiation work for such a wide class of functions, because you can always reduce the complexity. The absence of an equivalent for integration is what makes integration such a world of technique and tricks.

The key point I speak of, therefore, is that hardly any functions can be integrated! Given a function of any complexity, the chances of its antiderivative being an elementary function are very small.

This is deeply contrary to the expectations you build when learning integration - but that's because the lessons are focusing on functions you can integrate, which fortunately overlap closely with the sorts of elementary functions you'd have learned at that stage: trig, exp, polynomials, inverses. They don't focus on the absence of techniques on non-integrable functions, because there's not much to say, and that leaves the impression that having an elementary antiderivative is the norm. And when you think about it, the key technique in integration is spotting how to turn what you've got into the result of a differentiation, so you can run it backwards.

Fortunately, many of the functions that are integrable are common and useful, so it's by no means a lost battle. And the mine of analytical tricks is pretty deep. And when that runs out, there are approximate and numerical methods - Taylor series, Simpsons Rule and the like, or, as we say nowadays "computers" - for solving anything definite. Or we just give the result a nice name (eg erf) and leave it at that.

Oblig. xkcd: https://xkcd.com/2117/

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The chain rule for integration is basically $u$-substitution.

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    $\begingroup$ It will be a good answer with an example. $\endgroup$ – Basj Feb 1 '16 at 14:38
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    $\begingroup$ Though that's like saying that inegration by parts is the product rule of integration ... $\endgroup$ – Hagen von Eitzen Feb 2 '16 at 14:05
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    $\begingroup$ What is so special about $u$? $\endgroup$ – Carsten S Feb 2 '16 at 16:53
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    $\begingroup$ integration by parts is the product rule of integration, though. $\frac{d}{dx}[f(x)\cdot g(x)] = (f(x)\cdot g'(x)) \cdot (f'(x)\cdot g(x))$ means it follows immediately that $\int d[f(x)\cdot g(x)]=\int (f(x)\cdot g'(x))dx \cdot \int (f'(x)\cdot g(x))dx $ and therefore $\int (f(x)\cdot g'(x))dx = [f(x)\cdot g(x)] - \int (f'(x)\cdot g(x))dx$ $\endgroup$ – sig_seg_v Feb 2 '16 at 19:45
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    $\begingroup$ The name "u-substitution" seems to be widely used in US colleges, but is not a very useful name in general. :) $\endgroup$ – Pedro Tamaroff Feb 21 '16 at 14:02
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For calculating derivatives, we use the chain rule by multiplying by one.

$$\frac{dy}{dx}=\frac{dy}{dx}\cdot\frac{du}{du}=\frac{dy}{du}\cdot \frac{du}{dx}$$

Similarly, when integrating with the substitution rule, we also multiply by one. Here is a specific example.

$$\begin{array}{lll} \displaystyle\int_{x=0}^{x=2}xe^{x^2}dx &=& \displaystyle\int_{x=0}^{x=2}xe^{x^2}\color{red}{dx}\cdot\frac{\frac{dx^2}{\color{red}{dx}}}{\frac{dx^2}{dx}}\\ &=&\displaystyle\int_{x=0}^{x=2}\frac{xe^{x^2}\color{red}{dx}\cdot\frac{dx^2}{\color{red}{dx}}}{\frac{dx^2}{dx}}\\ &=&\displaystyle\int_{x=0}^{x=2}\frac{xe^{x^2}dx^2}{2x}\\ &=&\displaystyle\int_{x=0}^{x=2}\frac{e^{x^2}dx^2}{2}\\ &=&\displaystyle\int_{u=0}^{u=4}\frac{e^{u}du}{2}\\ \end{array}$$ Where $u=x^2$.

Note that the numerator of $\frac{\frac{dx^2}{dx}}{\frac{dx^2}{dx}}$ is interpreted as a ratio of differentials, whereas the denominator is interpreted as a derivative (function).

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  • $\begingroup$ so to sum up: We cannot solve the integral of 2 or more functions if the functions are not related together (ie. one is the derivative of the other). Same with quotients. And we use substitution for that. As for complex functions, can we find the derivative of any complex function? like sin(2x^(3x+2))? We use substitution for that again? $\endgroup$ – ergon Feb 1 '16 at 15:45
  • $\begingroup$ We use integration by parts only to solve a product of functions that they are not otherwise related (ie. one derivative of the other) ? Standard books and websites do not describe well when we use each rule. $\endgroup$ – ergon Feb 1 '16 at 16:04
  • $\begingroup$ Substitution is used when the integrated cotains "crap" that is easily canceled by dividing by the derivative of the substitution. In general, every expression composed of elementary functions/operators can be diffrentiate, but not the same can be said of integrals....consider $$\int e^{x^{2}}dx$$ $\endgroup$ – John Joy Feb 1 '16 at 23:38
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    $\begingroup$ The reason that standard books do not describe well when to use each rule is that you're supposed to do the exercises and figure it out for yourself. While you may make a few guidelines, experience is the best teacher, at least as far as applying integration techniques go. $\endgroup$ – John Joy Feb 1 '16 at 23:42
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It's possible by generalising Faa Di Bruno's formula to fractional derivatives then you can make the order of differentiation negative to obtain a series for for the n'th integral of f(g(x)).

Here's a paper detailing the fractional chain rule:

Fractional derivative of composite functions: exact results and physical applications,by Gavriil Shchedrin, Nathanael C. Smith, Anastasia Gladkina, Lincoln D. Carr

https://arxiv.org/abs/1803.05018

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Consider the functions z(y) and y(x). I am showing an example of a chain rule style formula to calculate $I(x) = \int dx z(y(x)),$ where z(y) can be triply integrated over dy, and where $y(x)=\sqrt{x}$ or $y(x)=x.$

To construct a formula for I(x), first define F(y) as the triple integration of z(y) over dy, that is $F(y) = \iiint dy dy dy z(y).$ Next evaluate F(y) for y(x), that is define $G(x) = F(y(x)).$ Differentiate G(x) twice over dx and then divide by $(dy/dx)^3,$ yielding $I(x) = \int dx z(y(x)) = G''(x) / y'^3.$

Consider an example calculation of I(x) where $z = y^3.$
First let $y(x)=\sqrt{x},$ so $dy/dx = (1/2) {x^{-1/2}}.$
Then $F(y) = y^6 / 120 + ay^2/2 + by + c,$ which yields $G(x) = F(y(x)) = x^3 /120 + ax/2 + bx^{1/2} + c,$ and then $G''(x) = x/20 - (1/4)bx^{-3/2},$ so that $I(x) = y'^{-3} G''(x) = 8 x^{3/2} [ x/20 - (1/4)bx^{-3/2} ]= (2/5)x^{5/2} - 2b.$ Directly integrating for $y = x^{1/2}$ and $z = y^3$ yields $I(x) = \int dx z(y(x)) = \int dx y^3 = \int dx x^{3/2} = (2/5) [x^{5/2}] + constant.$

Next for this same example z = $y^3$ let y = x. Then $F(y) = y^6/120 + ay^2/2 + by + c,$ which yields
$G(x) = F(y(x)) = x^6/120 + ax^2/2 + bx + c,$ so that
$I(x) = y'^{-3}G''(x) = 1^{-3} [x^4/4 + a] = x^4/4 + a.$
Directly integrating yields $I(x) = \int dx z(y(x)) = \int dx y^3 = \int dx x^3 = x^4/4 + constant.$

This demonstrates that the direct and chain rule methods agree with each other to within a constant for $y(x)=x$ and $y(x)=\sqrt{x}$ for the specific function $z(y) = {y^3}.$ This agreement should work for any function z(y) where $y(x)=x$ or $y(x)=\sqrt{x}.$

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  • $\begingroup$ It would be interesting to see if the above-mentioned Faa Di Bruno's formula generalized to fractional derivatives could be used to calculate this formula for I(x). $\endgroup$ – rich0962 Dec 27 '20 at 0:27
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If $g$ is invertible, then the integral $$ \int f(g(x)) \, dx $$ can be attacked using integration by substitution. Set $u=g(x)$, meaning that $x=g^{-1}(u)$ and $dx=(g^{-1})'(u)du$. The integral becomes $$ \int f(u)(g^{-1})'(u) \, du \, . $$ There are instances where the transformed integral is easier to solve than the original one, but there is no guarantee of this, and obviously we still don't have a 'rule' that tells us that $$ \int f(g(x)) \, dx = F(x) + C $$ for some function $F$.


For an example of when this technique might be useful, consider $$ \int \sqrt{1-x^2} \, dx \, . $$ This integral looks impossibly hard to solve until you make the substitution $\theta=\arcsin x$. Note that the inner function $g$ is in disguise: $$ \int \sqrt{1-x^2} \, dx = \int \sqrt{1-\sin^2(\arcsin x)} \, dx \, . $$ (In practice, we tend to say 'let $x=\sin \theta$' and proceed from there, but this is not quite correct because $x=\sin \theta$ does not imply $\arcsin x =\theta$.)

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There is no general chain rule for integration known.

The goal of indefinite integration is to get known antiderivatives and/or known integrals.

To get chain rules for integration, one can take differentiation rules that result in derivatives that contain a composition and integrate this rules once or multiple times and rearrange then. For some kinds of integrands, this special chain rules of integration could give known antiderivatives and/or known integrals.

From a chain rule, we expect that the left-hand side of the equation is $\int f(g(x))dx$.

For linear $g(x)$, the commonly known substitution rule

$$\int f(g(x))\cdot g'(x)dx=\int f(t)dt;\ t=g(x)$$

becomes a chain rule.

Further chain rules are written e.g. in
Will, J.: Product rule, quotient rule, reciprocal rule, chain rule and inverse rule for integration. May 2017:

Let
$c$ be an integration constant,
$\gamma$ be the compositional inverse function of function $g$,
$F(g(x))=\int f(t)dt+c;\ t=g(x)$.

$$\int f(g(x))dx=\int f(t)\gamma'(t)dt;\ t=g(x)$$

$$\int f(g(x))dx=xf(g(x))-\int f'(t)\gamma(t)dt;\ t=g(x)$$

$$\int f(g(x))dx=\left(\frac{d}{dx}F(g(x))\right)\int\frac{1}{g'(x)}dx-\int \left(\frac{d^{2}}{dx^{2}}F(g(x))\right)\int\frac{1}{g'(x)}dx\ dx$$

$$\int f(g(x))dx=\frac{F(g(x))}{g'(x)}+\int F(g(x))\frac{g''(x)}{g'(x)^{2}}dx$$

The complexity of the integrands on the right-hand side of the equations suggests that these integration rules will be useful only for comparatively few functions. For linear g(x) however the integrand on the right-hand side of the last equation simplifies advantageously to zero. But this is already the substitution rule above.

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