When the inside of a bracket is a linear expression ax + b, integrate as if it were a single variable, then divide by a. For example, ∫(3x − 2)4 dx = (3x − 2)5/15 + c.
This builds on integrating a power with the correct constant. Check the current 0606 syllabus page for the exact functions you need, because the same idea also covers forms such as eax+b.
Why divide by the inner coefficient?
Differentiating (3x − 2)5 with the chain rule gives 5(3x − 2)4 × 3. The extra 3 is the derivative of the inside.
So to cancel both the 5 from the power and the 3 from the inside, you divide by 5 × 3 = 15. The integral is a reverse chain rule.
How do you do it, step by step?
- Identify the inside ax + b and the outside rule (a power, e…, sin or cos).
- Integrate the outside as if the inside were just x.
- Divide by a, the coefficient of x in the inside.
- Add + c, then differentiate mentally to check.
Worked example
Find (a) ∫(3x − 2)4 dx, (b) ∫e2x+1 dx, and (c) ∫6x(x² + 1)³ dx.
(a) Step 1, inside and outside: the inside is 3x − 2, so a = 3. The outside is a power of 4.
Step 2, integrate and divide: (3x − 2)5/5, then divide by 3.
(3x − 2)5/15 + c
(b) The integral of eu is eu, and a = 2, so:
½e2x+1 + c
(c) The inside x² + 1 has derivative 2x, and the bracket is multiplied by 6x = 3 × 2x. Treat (x² + 1) as u, so the integral of 2x(x² + 1)³ is (x² + 1)⁴/4. Multiply by 3:
3(x² + 1)⁴/4 + c
Check of (c): differentiating gives 3 × 4(x² + 1)³ × 2x / 4 = 6x(x² + 1)³, which is the original.
Part (c) is a reverse chain rule where the inside is not linear. It only works because 6x is a constant multiple of the derivative of x² + 1. Check your syllabus for whether this form is required.
The mistake to watch for
A common slip is to ignore the inner coefficient.
Mistaken answer: ∫(3x − 2)4 dx = (3x − 2)5/5 + c
Differentiating this gives 3(3x − 2)4, which is three times too big.
The correction is to divide by 3 as well, giving (3x − 2)5/15. Always differentiate your answer once. If you get a multiple of the original, the missing factor is the inner coefficient.
Check yourself
Try these on paper, then open each answer.
1. Find ∫(2x + 5)³ dx.
Show answer
Power: (2x + 5)⁴/4. Divide by the inner coefficient 2, so the divisor is 8. (2x + 5)⁴/8 + c
2. Find ∫sin(3x) dx.
Show answer
The integral of sin u is −cos u. Divide by 3. −cos(3x)/3 + c
3. Find ∫10e5x−1 dx.
Show answer
The integral of e5x−1 is e5x−1/5. Multiply by 10. 2e5x−1 + c
Where this leads next
Next, put limits on these integrals in evaluating a definite integral with correct limits, and keep checking an antiderivative by differentiation as a habit. The non-calculator working trainer and the quadratic structure explorer help with the algebra behind the brackets.
If reverse chain rule questions are where you stall, our teachers can work on them in online one-to-one Additional Mathematics tuition.