The reverse chain rule
The reverse chain rule is a way to integrate a function by spotting that it came from the chain rule. If the thing you are integrating is a function of something, multiplied by the derivative of that something, you can write the answer down directly: ∫ f′(x) [f(x)]ⁿ dx = [f(x)]ⁿ⁺¹ ÷ (n + 1) + c. This is also called integration by inspection. The method is to guess the answer, differentiate it to check, and adjust the number in front.
Why the reverse chain rule works
Differentiating with the chain rule multiplies by the derivative of the inside. For example, the derivative of (x² + 1)⁵ is 5(x² + 1)⁴ × 2x = 10x(x² + 1)⁴. Integration undoes differentiation, so the integral of 10x(x² + 1)⁴ must be (x² + 1)⁵ + c.
So whenever an integral contains a function of g(x) multiplied by g′(x), or a number times g′(x), it is the result of a chain rule, and you can work backwards.
The reverse chain rule patterns
| Integral | Result | Example |
|---|---|---|
| ∫ (ax + b)ⁿ dx | (ax + b)ⁿ⁺¹ ÷ [a(n + 1)] + c | ∫ (3x + 2)⁵ dx = (3x + 2)⁶ ÷ 18 + c |
| ∫ f′(x) [f(x)]ⁿ dx | [f(x)]ⁿ⁺¹ ÷ (n + 1) + c, for n ≠ −1 | ∫ 2x(x² + 1)⁴ dx = (x² + 1)⁵ ÷ 5 + c |
| ∫ f′(x) ÷ f(x) dx | ln f(x) + c, using the size of f(x) | ∫ 2x ÷ (x² + 5) dx = ln(x² + 5) + c |
| ∫ f′(x) e^f(x) dx | e^f(x) + c | ∫ 3x² e^(x³) dx = e^(x³) + c |
| ∫ cos(ax + b) dx | sin(ax + b) ÷ a + c | ∫ cos 4x dx = ¼ sin 4x + c |
| ∫ sin(ax + b) dx | −cos(ax + b) ÷ a + c | ∫ sin 2x dx = −½ cos 2x + c |
In every case the derivative of the inside function is either already there or there apart from a constant you can fix.
How to use the reverse chain rule
- Spot the inside function, such as the bracket, the power of e or the bottom of a fraction.
- Check its derivative is there, apart from a number. If it is not, the reverse chain rule will not work and you need substitution or another method.
- Guess the answer: raise the bracket's power by one, or write ln of the denominator, or keep the e part.
- Differentiate your guess and compare it with what you were integrating.
- Adjust the constant in front so that they match, and add + c.
Example 1: a linear bracket
Find ∫ (3x + 2)⁵ dx.
Guess (3x + 2)⁶. Differentiating gives 6(3x + 2)⁵ × 3 = 18(3x + 2)⁵, which is 18 times too big. So divide by 18:
Example 2: the derivative is there, apart from a number
Find ∫ x(x² + 1)⁴ dx.
The inside is x² + 1, whose derivative is 2x. The integral has x, which is 2x apart from a factor of ½, so the reverse chain rule works. Guess (x² + 1)⁵: its derivative is 5(x² + 1)⁴ × 2x = 10x(x² + 1)⁴. That is 10 times too big:
Example 3: an exponential
Find ∫ 6x² e^(x³) dx.
The derivative of x³ is 3x², and 6x² is twice that. Guess e^(x³): its derivative is 3x² e^(x³), so you need twice it:
Example 4: a fraction that gives a log
Find ∫ 4x ÷ (x² + 5) dx.
The derivative of the bottom, x² + 5, is 2x, and the top is twice that. So the answer is twice ln of the bottom:
x² + 5 is always positive, so no modulus signs are needed here. In general, ∫ f′(x) ÷ f(x) dx = ln|f(x)| + c.
Example 5: a power of a trig function
Find ∫ sin x cos³x dx.
cos³x is (cos x)³, and the derivative of cos x is −sin x, which is there apart from the sign. Guess cos⁴x: its derivative is 4cos³x × (−sin x) = −4 sin x cos³x. So multiply by −¼:
Example 6: a definite integral
Evaluate the integral of x√(1 + x²) from 0 to 1.
Write it as x(1 + x²)^(1/2). Guess (1 + x²)^(3/2): its derivative is (3/2)(1 + x²)^(1/2) × 2x = 3x(1 + x²)^(1/2). So the integral is (1 + x²)^(3/2) ÷ 3.
That is about 0.609. Check it with the Integral Calculator.
This practice from the Calculus course integrates functions term by term. Use it to warm up the basic rules that the reverse chain rule builds on, then press Mark to check your answer in the course.
Integrate a random polynomial, remembering the constant of integration.
Press Mark to have your answer marked in the Calculus course.
When the reverse chain rule does not work
The method needs the derivative of the inside function, up to a constant. If anything else involving x is missing or extra, it fails.
- ∫ (x² + 1)⁴ dx: the 2x is missing, and you cannot put in an x with a constant. Expand the bracket instead.
- ∫ x²(x² + 1)⁴ dx: x² is not a constant multiple of 2x. Use substitution or expand.
- ∫ e^(x²) dx: there is no x outside, and this integral has no answer in simple functions at all.
For harder cases, integration by substitution is the general method that the reverse chain rule is a shortcut for.
Common mistakes
- Forgetting to divide by the inside derivative. ∫ (3x + 2)⁵ dx is (3x + 2)⁶ ÷ 18, not ÷ 6.
- Applying it when the derivative is missing. Check that the inside derivative is really there, up to a number.
- Not checking by differentiating. It takes seconds and catches nearly every slip.
- Losing a minus sign with trig. The derivative of cos x is −sin x, and the integral of sin x is −cos x.
- Forgetting + c on an indefinite integral.
- Forgetting the modulus in ln|f(x)| when f(x) could be negative.
Is the reverse chain rule in the formula booklet?
Not as a rule. Of the A-level Maths booklets checked in November 2026, AQA's gives ∫ f′(x) ÷ f(x) dx = ln|f(x)| + c, one of the reverse chain rule patterns, and Edexcel's does not. Both give integration by parts. The other patterns are not printed in either, so you need to recognise them yourself.
Reverse chain rule questions with answers
1. Find ∫ (2x − 1)⁶ dx.
(2x − 1)⁷ ÷ 14 + c
2. Find ∫ x²(x³ + 2)⁴ dx.
(x³ + 2)⁵ ÷ 15 + c
3. Find ∫ cos x ÷ sin x dx.
ln|sin x| + c
4. Find ∫ (2x + 3) ÷ (x² + 3x) dx.
ln|x² + 3x| + c
5. Evaluate the integral of sin x cos²x from 0 to π/2.
[−cos³x ÷ 3] from 0 to π/2 = 0 − (−⅓) = 1/3
Practise integration in a short timed paper
A-level Maths: Calculus gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across differentiation and integration. In the US, the same practice opens as AP Calculus AB, where the method is usually called u-substitution.
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