Teacher page: indefinite integrals and the reverse chain rule (AA 5.10)

The one thing to do with the figure

Establish differentiating back as the default.

The checker compares the proposed answer, differentiated numerically, with the integrand. Use it twice and then point out that they can do the same in their heads in five seconds.

This is the habit that turns integration from guesswork into something a student can be certain about, and it costs no extra time in an exam.

n = −1 is now in scope

At 5.5 the index had to be a whole number other than −1, so 1/x was simply not integrated. Here it is, and it gives ln|x|.

Expect someone to try "add one to the power" on 1/x and get division by zero. That failure is a good thirty seconds: it is exactly why the logarithm case exists.

The answers

Linear insides	∫cos(2x+3) = ½sin(2x+3) + C; ∫1/(3x+2) = ⅓ln\|3x+2\| + C.
Reverse chain	∫2x(x²+1)⁴ = (x²+1)⁵/5 + C; ∫4x sin(x²) = −2cos(x²) + C.
1. ∫(1/x)dx	B, ln\|x\| + C.
2. ∫cos(2x+3)dx	B, ½sin(2x+3) + C.
3. By inspection	B, ∫2x(x²+1)⁴ dx.

Where the marks go

1 markIdentifying the inner function, ideally with u and du written out.

1 markThe dividing constant.

1 markThe + C and the modulus in the logarithm.

What each wrong answer tells you

They give	What it means
Multiplying by the inside derivative	Running the chain rule forwards. The factor goes on the bottom, not the top.
x⁰/0 for ∫1/x	The trap, and a productive one. Let it happen before you give the logarithm.
Attempting ∫sin(x²)	They think any inside works. Be explicit: without the factor there is no elementary answer and none is expected.
Missing + C	Routine, and routinely a mark.

Other things they will say

"Can I always substitute?" Only when the inside derivative is there up to a constant. A constant you can fix; a missing variable you cannot.

"Why the modulus?" The argument of a logarithm must be positive and the expression may not be. It costs nothing to write.

"Does the calculator do it?" For definite integrals yes, and that is a sensible check. For indefinite ones you need the working.

A possible order

	What is happening
1	Differentiate three chain rule examples, then show the answers and ask them to get back.
2	The pattern, naming the inside every time.
3	The checker. Make it a habit.
4	Linear insides, then the non-linear case where the factor must be present.
5	The logarithm case, and questions 1 to 3.
6	A mixed set including some that cannot be done.

Two things not to say

Do not set an exercise where everything works. Deciding whether the pattern is there is part of the skill.

Do not skip the formal u and du for strong students. They will be stuck the first time the substitution is not obvious.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

Spot the derivativeEvaluate the integral of 2xe^x² from 0 to 1.
With u = x², it is the integral of e^u from 0 to 1 = e − 1 = 1.718.
A logarithmEvaluate the integral of 1/x from 1 to 3.
ln 3 = 1.099.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

Change the limitsEvaluate the integral of 3x²(x³ + 1)² from 0 to 1.
With u = x³ + 1, u runs 1 to 2 and the integral of u² is (8 − 1)/3 = 7/3 = 2.333.
Pick the substitutionFor the integral of x/(x² + 4), state the substitution and the answer.
u = x² + 4, since du = 2x dx is present up to a factor of 2. The result is (1/2)ln|x² + 4| + c.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

Keep the modulusState the integral of 1/x and explain why the modulus matters for AA.
ln|x| + c. Without it the antiderivative is undefined for negative x, so it cannot be used on an interval left of the origin. AA assesses the integral of 1/x at Standard Level and the modulus is sometimes required explicitly.
When substitution will not workExplain why u = x² fails for the integral of e^x² alone.
du = 2x dx needs an x outside, and there is none to supply it. In fact that integral has no elementary antiderivative at all, which is why the question is always set with the 2x present.

Practicalities

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