Inverse Chain Rule Method: Reverse Chain Rule Integration

What Is The Inverse Chain Rule Method?

The inverse chain rule method is a technique for integrating a product of the form "a function of an inside function, times the derivative of that inside function", by recognising it as the output of the chain rule and writing down what was differentiated. A question posted on Vedantu asks it directly: is there an inverse chain rule for integration? There is, and in its general form it reads:

$$\int f'(g(x)),g'(x),dx = f(g(x)) + C$$

Textbooks often phrase the same pattern as "$f'(x)$ times $g(f(x))$"; this page names the inside function $g$ instead. Here $g(x)$ is the inside function, $g'(x)$ is its derivative, $f$ is any function whose derivative $f'$ you know, and $C$ is the constant of integration. The rule holds on any interval where $g$ is differentiable with a continuous derivative and $f'$ is continuous on the values $g$ takes.

The most used special case is the power version, for any constant $n \neq -1$:

$$\int g'(x),\left[g(x)\right]^n,dx = \frac{\left[g(x)\right]^{n+1}}{n+1} + C$$

For $n = -1$ the power formula would divide by zero, and the logarithm takes over instead: $\int \frac{g'(x)}{g(x)},dx = \ln\left|g(x)\right| + C$. Every answer on this page is an indefinite integral, so every answer carries $+ C$.

Is The Inverse Chain Rule The Same As The Inverse Function Rule?

No. The inverse chain rule method is an integration technique, while the inverse function rule is a differentiation formula, and the similar names cause real confusion in searches and in revision notes.

The inverse function rule answers a different question: if you know the slope of $f$, what is the slope of $f^{-1}$? That belongs with inverse functions and differentiation, and it never produces an integral.

The method also travels under other names, which is why a thread on The Student Room asks for the difference between the reverse chain rule and inspection. Integration by inspection (or by recognition) is the broader habit: guess an antiderivative by eye, then differentiate to check. The reverse chain rule is inspection aimed at one shape, the chain-rule product. The name "reverse chain rule" is most common in British A-level teaching, while American textbooks usually go straight to u-substitution for the same move.

Why Does The Inverse Chain Rule Method Work?

The method works because integration undoes differentiation, so every derivative rule read backwards is an integration rule. The chain rule says that for a composite function:

$$\frac{d}{dx}\Big[f(g(x))\Big] = f'(g(x)),g'(x)$$

By definition, an antiderivative of the right-hand side is anything whose derivative is the right-hand side. The left-hand side already tells you one: $f(g(x))$. Adding any constant keeps the derivative the same, so:

$$\int f'(g(x)),g'(x),dx = f(g(x)) + C$$

That answers The Student Room question about differentiation and integration using the chain rule and the reverse chain rule: they are one identity read in two directions. The forward direction multiplies by the derivative of the inside; the backward direction needs that derivative to be present already, so it can be absorbed.

There is a picture behind it too. Over an interval $[a, b]$, the substitution $u = g(x)$ stretches the horizontal axis, and the factor $g'(x),dx$ is the width of a thin strip after the stretch. So the area under $f'(g(x)),g'(x)$ from $a$ to $b$ equals the area under $f'(u)$ from $g(a)$ to $g(b)$. For instance, the area under $2x\sin(x^2)$ from $0$ to $\sqrt{\pi}$ and the area under $\sin u$ from $0$ to $\pi$ are both exactly $2$.

How Do You Spot An Inside Function And Its Derivative?

You spot it by naming the inside function first, then checking whether its derivative sits elsewhere in the integrand, up to a constant multiple. A thread on The Student Room puts the real worry plainly: when do I use the reverse chain rule? Use it when this three-question check passes:

1. Is there an inside function? Look for something inside a bracket raised to a power, inside a root, in an exponent, inside a trig function, or in a denominator.

2. Is its derivative also present? Differentiate the inside function and look for the result as a factor.

3. Is the match exact up to a constant? A missing $2$ or $-\frac{1}{3}$ is fine. A missing $x$ is not.

If all three answers are yes, read the antiderivative off this table (the first row needs $n \neq -1$).

The habit that makes this fast is saying the inside function's derivative out loud before writing anything. Students who skip that step tend to see "a bracket to a power" and apply the power rule to the bracket alone, which is right only when the inside derivative is $1$.

How Do You Integrate $(ax+b)^n$?

You integrate $(ax+b)^n$ by raising the power by one, dividing by the new power, and dividing by $a$, the derivative of the inside:

$$\int (ax+b)^n,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C, \qquad a \neq 0,\ n \neq -1$$

The extra $\frac{1}{a}$ is there because differentiating $(ax+b)^{n+1}$ by the chain rule produces a factor of $a$, and the answer must cancel it. The same correction appears for every linear inside function:

• Logarithm case: $\int \frac{1}{ax+b},dx = \frac{1}{a}\ln\left|ax+b\right| + C$, the $n = -1$ case, built on the natural logarithm.

• Exponential case: $\int e^{3x+1},dx = \frac{1}{3}e^{3x+1} + C$.

• Trig case: $\int \sin(2x+1),dx = -\frac{1}{2}\cos(2x+1) + C$.

Linear inside functions are the friendliest case, since their derivative is a constant and a constant can always be fixed. Check any answer by differentiating it: $\frac{d}{dx}\left[\frac{(3x-2)^6}{18}\right] = \frac{6 \cdot 3,(3x-2)^5}{18} = (3x-2)^5$.

How Do You Adjust The Constants?

You adjust a constant by multiplying inside the integral and dividing outside by the same number, which leaves the value unchanged. This is the constant multiple rule at work. Take $\int x,(x^2+5)^4,dx$:

$$\frac{d}{dx}\left(x^2+5\right) = 2x, \text{ but only } x \text{ is present}$$

$$\int x,(x^2+5)^4,dx = \frac{1}{2}\int 2x,(x^2+5)^4,dx$$

$$= \frac{1}{2}\cdot\frac{(x^2+5)^5}{5} + C = \frac{(x^2+5)^5}{10} + C$$

The repair only works for constants, because a constant can slide in and out of an integral sign and $x$ cannot. Faced with $\int (x^2+5)^4,dx$, there is no $x$ to complete, and dividing by $2x$ outside the integral is not allowed. That integral needs expanding instead.

Only constants can be adjusted. If the missing piece contains $x$, the reverse chain rule does not apply, and another method is needed.

How Does The Inverse Chain Rule Relate To U-Substitution?

The inverse chain rule method and u-substitution are the same theorem, written two ways. A Quora question asks why u-substitution for integration is the reverse of the chain rule, and the answer is that u-substitution simply names the inside function $u = g(x)$ and records $du = g'(x),dx$, while the reverse chain rule does both steps in your head.

Here is one integral done both ways:

So when a student on The Student Room asks whether they can just use substitution instead of struggling with the reverse chain rule, the answer is yes, always. Substitution is slower but makes every step visible, and the full method, including choosing $u$ when the pattern is hidden, lives on the integration by substitution page. Once the pattern is familiar, the mental version saves lines; when it is not, writing $u$ down is the safer choice.

For definite integrals, the mental version keeps the original limits, because the answer is written back in $x$ before you substitute the limits. If you switch to $u$, the limits must switch to $g(a)$ and $g(b)$ as well.

Examples Of The Inverse Chain Rule Method

The six examples below move from a linear inside function to a definite integral.

Example 1

Find $\int (3x-2)^5,dx$.

Inside function $3x - 2$, derivative $3$, a constant.

$$\int (3x-2)^5,dx = \frac{(3x-2)^6}{6 \cdot 3} + C$$

$$= \frac{(3x-2)^6}{18} + C$$

Final answer: $\frac{(3x-2)^6}{18} + C$.

Example 2

Find $\int 6x^2,(x^3+1)^4,dx$.

Inside function $x^3 + 1$, derivative $3x^2$. The integrand has $6x^2 = 2 \cdot 3x^2$.

$$\int 6x^2,(x^3+1)^4,dx = 2\int 3x^2,(x^3+1)^4,dx$$

$$= 2 \cdot \frac{(x^3+1)^5}{5} + C$$

$$= \frac{2(x^3+1)^5}{5} + C$$

Final answer: $\frac{2(x^3+1)^5}{5} + C$.

Example 3

Find $\int \cos^2 x,dx$.

This one comes from a Quora question: why can't we integrate $(\cos x)^2$ by the reverse chain rule? It looks like a bracket squared, so the tempting move is the power pattern:

$$\int (\cos x)^2,dx \overset{\text{tempting}}{\longrightarrow} \frac{\cos^3 x}{3} + C$$

Differentiate that guess to test it:

$$\frac{d}{dx}\left[\frac{\cos^3 x}{3}\right] = \cos^2 x \cdot (-\sin x)$$

The result has an extra factor of $-\sin x$, so the guess is wrong. The power pattern needs the inside derivative, $-\sin x$, to be present in the integrand, and here it is not. Dividing by $-\sin x$ to fix it is not allowed either, because $-\sin x$ is not a constant.

The rescue is to rewrite the integrand first with the double-angle identity, so that only linear inside functions remain:

$$\cos^2 x = \frac{1 + \cos 2x}{2}$$

$$\int \cos^2 x,dx = \int \frac{1}{2},dx + \frac{1}{2}\int \cos 2x,dx$$

$$= \frac{x}{2} + \frac{1}{2} \cdot \frac{\sin 2x}{2} + C$$

The reverse chain rule did still help, on $\cos 2x$, whose inside derivative is the constant $2$.

Final answer: $\frac{x}{2} + \frac{\sin 2x}{4} + C$.

Example 4

Find $\int \tan x,dx$.

Write $\tan x = \frac{\sin x}{\cos x}$. The inside function in the denominator is $\cos x$, with derivative $-\sin x$.

$$\int \frac{\sin x}{\cos x},dx = -\int \frac{-\sin x}{\cos x},dx$$

$$= -\ln\left|\cos x\right| + C$$

Final answer: $-\ln\left|\cos x\right| + C$, which is the same as $\ln\left|\sec x\right| + C$.

Example 5

Find $\int \frac{(\ln x)^3}{x},dx$ for $x > 0$.

Inside function $\ln x$, derivative $\frac{1}{x}$, present exactly.

$$\int \frac{1}{x},(\ln x)^3,dx = \frac{(\ln x)^4}{4} + C$$

Final answer: $\frac{(\ln x)^4}{4} + C$.

Example 6

Evaluate $\int_{0}^{2} x\sqrt{x^2+5},dx$.

Inside function $x^2 + 5$, derivative $2x$; only $x$ is present, so adjust by $\frac{1}{2}$. The power is $\frac{1}{2}$.

$$\int x,(x^2+5)^{1/2},dx = \frac{1}{2} \cdot \frac{(x^2+5)^{3/2}}{3/2} + C = \frac{(x^2+5)^{3/2}}{3} + C$$

$$\int_{0}^{2} x\sqrt{x^2+5},dx = \frac{9^{3/2} - 5^{3/2}}{3}$$

$$= \frac{27 - 5\sqrt{5}}{3} = 9 - \frac{5\sqrt{5}}{3}$$

Final answer: $9 - \frac{5\sqrt{5}}{3} \approx 5.2732$.

Why Does Integration Need A Reverse Chain Rule?

Integration needs a reverse chain rule because, unlike differentiation, it has no general rule for composite functions. "Every derivative can be computed; not every integral can" is the real gap the method fills:

• Differentiation always succeeds. Product, quotient and chain rules together differentiate any formula built from standard functions.

• Integration often fails. The integral $\int e^{x^2},dx$ has no antiderivative in elementary functions at all; its value on $[0, 1]$ is about $1.4627$, found numerically, and it is a nonelementary integral.

• The chain-rule footprint is what rescues it. Add one factor of $x$ and everything changes: $\int x,e^{x^2},dx = \frac{1}{2}e^{x^2} + C$, because the inside derivative is now present.

So the method is less a trick than a test: it tells you whether an integrand is one of the composites that integrate in closed form. The notation makes this feel natural. Gottfried Leibniz first wrote the integral sign with $dx$ in a 1675 manuscript, and because $du = g'(x),dx$ looks like ordinary algebra in his notation, the substitution reads almost like cancelling.

Where Does The Inverse Chain Rule Show Up Beyond The Exam?

The inverse chain rule appears wherever a rate has an exponential or power form with a constant inside, which covers much of physics and biology.

• Decay and clearance. A drug cleared at a rate proportional to its concentration follows $e^{-kt}$, and the total exposure is $\int_{0}^{\infty} e^{-kt},dt = \frac{1}{k}$, a linear-inside-function integral.

• Circuits. The current in a discharging capacitor decays like $e^{-t/RC}$; the charge that flows is its integral, with the $\frac{1}{a}$ correction giving the factor $RC$.

• Probability. The bell-curve shape $x,e^{-x^2/2}$ integrates to $-e^{-x^2/2} + C$ by the reverse chain rule, an integral that appears whenever averages involving the normal distribution are computed.

• Work and energy. Forces that depend on a power of distance integrate with the power pattern; the fundamental theorem of calculus then turns the antiderivative into a number.

Tripping Points To Avoid

The inverse chain rule method fails in four predictable places, and each one can be caught by differentiating the answer.

Mistake 1: Forgetting To Divide By The Inside Derivative

Where it slips in:

Linear inside functions such as $\cos(4x)$ or $(5x+1)^7$.

Don't do this:

Writing $\int \cos(4x),dx = \sin(4x) + C$. Differentiating gives $4\cos(4x)$, four times too big.

The correct way:

Divide by the inside derivative: $\int \cos(4x),dx = \frac{\sin(4x)}{4} + C$. The rusher, who knows $\int \cos x,dx = \sin x$ by heart, writes the familiar answer and moves on; the differentiation check takes five seconds and catches it every time.

Mistake 2: Adjusting By A Variable

Where it slips in:

Integrands where the inside derivative contains $x$ but the integrand does not, such as $\int (x^2+1)^3,dx$.

Don't do this:

Writing $\frac{(x^2+1)^4}{4 \cdot 2x} + C$. Differentiating it back gives $(x^2+1)^3$ minus an extra term $\frac{(x^2+1)^4}{8x^2}$, so it is wrong.

The correct way:

Only constants move through an integral sign. Expand instead: $\int (x^2+1)^3,dx = \frac{x^7}{7} + \frac{3x^5}{5} + x^3 + x + C$. The memorizer who learned "divide by the derivative of the bracket" as a rule, without the reason, makes exactly this move; the first instinct to treat $2x$ like the constant $a$ is the error to watch for.

Mistake 3: Using The Power Pattern When $n = -1$

Where it slips in:

Fractions such as $\frac{2x}{x^2+1}$, which are $g'(x)\left[g(x)\right]^{-1}$.

Don't do this:

Writing $\frac{(x^2+1)^0}{0}$, which divides by zero.

The correct way:

Use the logarithm row: $\int \frac{2x}{x^2+1},dx = \ln(x^2+1) + C$. No absolute value is needed here because $x^2 + 1$ is always positive.

Mistake 4: Keeping The Old Limits After Switching To $u$

Where it slips in:

Definite integrals done with an explicit substitution.

Don't do this:

Changing $\int_{0}^{2} x\sqrt{x^2+5},dx$ into a $u$-integral but keeping the limits $0$ and $2$.

The correct way:

Either change the limits to $u = 5$ and $u = 9$, or return to $x$ before substituting $0$ and $2$. Both give $9 - \frac{5\sqrt{5}}{3}$, as in Example 6.

Practice Problems On The Inverse Chain Rule Method

Work each one, then check by differentiating your answer. Every answer below has been recomputed.

1. Find $\int (5x+1)^7,dx$.

2. Find $\int x^2,e^{x^3},dx$.

3. Find $\int \frac{\cos x}{1+\sin x},dx$.

4. Find $\int \sec^2(3x),dx$.

5. Find $\int \frac{x}{(x^2+4)^3},dx$.

6. Evaluate $\int_{0}^{\pi/2} \sin x\cos^3 x,dx$.

Answer to Question 1: $\frac{(5x+1)^8}{40} + C$.

Answer to Question 2: $\frac{1}{3}e^{x^3} + C$.

Answer to Question 3: $\ln(1+\sin x) + C$ (the absolute value is not needed, since $1 + \sin x \geq 0$).

Answer to Question 4: $\frac{\tan(3x)}{3} + C$.

Answer to Question 5: $-\frac{1}{4(x^2+4)^2} + C$.

Answer to Question 6: $\frac{1}{4}$.

The Short Version

• The inverse chain rule method integrates $f'(g(x)),g'(x)$ to $f(g(x)) + C$ by reading the chain rule backwards.

• It is not the inverse function rule, which differentiates $f^{-1}$.

• For linear inside functions, divide by $a$: $\int (ax+b)^n,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$ for $n \neq -1$.

• Only constants can be adjusted; a missing factor of $x$ means another method is needed.

• U-substitution is the same theorem with the inside function written as $u$.

Your Next Move

Work through Practice Problems 1 to 6; if Problem 2 or 5 trips you up, return to "How Do You Adjust The Constants?", and if Problem 3 does, reread the logarithm row of the spotting table.

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Read More

• Antiderivatives: what it means for one function to undo another's derivative.

• Basic integration formulas: the outside-function results the spotting table relies on.

• Methods of integration: where the reverse chain rule sits among parts, partial fractions and substitution.

• Integration of trigonometric functions: identities like the one that rescued Example 3.

• Table of integrals: a lookup for the standard forms after the inside function is handled.