01 · Find the missing coefficient
Find k if k e³ˣ⁻¹ has derivative 6e³ˣ⁻¹.
Hint
Differentiation produces 3k.
Worked solution
3k = 6, so k = 2.
Understand · explore · practise
Integrating functions of ax + b
Integrate powers, exponentials and trigonometric functions with linear arguments ax + b. Understand the 1/a factor, logarithmic exceptions, negative slopes and exact definite integrals.
Before you startStandard integrals and the chain rule for differentiation.
01 / Undo the inner derivative
The compensating factor is 1/a.
If F′(u) = f(u), then ∫f(ax + b) dx = F(ax + b)/a + C
This rule requires a ≠ 0, on a valid interval.
Differentiating F(ax + b) multiplies by a. Integration must cancel that extra factor. The constant b stays inside the argument but contributes no derivative.
Cancel the inner derivativeExplore
At x = 0, u = 4.
Target integrand f(u) = 16.
The unscaled candidate derivative is 32.
Multiply by 1/2 to obtain the required derivative 16.
A primitive is (ax + 4)³/(3a), plus C.
Unscaled primitive value: 21.33333333. Corrected primitive value: 10.66666667.
The model uses b = 4. Its choices keep u positive for the logarithm, and a nonzero. The multiplier identity remains true even when the target integrand is zero; then the displayed bars are not numerical ratios.
02 / Derive and check the factor
Differentiate the proposed antiderivative.
Integrate e²ˣ⁺⁴.Worked example
Try e²ˣ⁺⁴
Its derivative is 2e²ˣ⁺⁴: twice the target.
Multiply the candidate by 1/2
The two factors cancel on differentiation.
∫e²ˣ⁺⁴ dx = (1/2)e²ˣ⁺⁴ + C
The shift 4 remains in the exponent.
Watch: cancel the inner chain-rule factor
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Integrate a power of a linear expression
Use both the new exponent and the inner slope.
∫(ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/[a(n + 1)] + C
a ≠ 0 and n ≠ −1; check the real domain for fractional powers.
Find ∫(5 − 2x)⁴ dx.Worked example
a = −2 and n + 1 = 5
Keep the negative slope.
Divide the fifth power by (−2)×5
Both factors appear in the denominator.
−(5 − 2x)⁵/10 + C
Differentiation gives the original fourth power.
02 · A square root
Find ∫√(3x + 4) dx where 3x + 4 > 0.
Hint
n = 1/2, so a(n + 1) = 9/2.
Worked solution
(2/9)(3x + 4)³ᐟ² + C.
03 · A negative power
Find ∫(2x + 5)⁻³ dx.
Hint
Divide by 2×(−2).
Worked solution
−1/[4(2x + 5)²] + C, on an interval avoiding x = −5/2.
04 / Handle the reciprocal separately
The exponent −1 still needs a logarithm.
∫1/(ax + b) dx = (1/a)ln|ax + b| + C
a ≠ 0; the interval must avoid ax + b = 0.
Find ∫7/(4 − 3x) dx.Worked example
The denominator has derivative −3
The logarithm candidate brings an unwanted factor −3.
Multiply ln|4 − 3x| by −7/3
Its derivative becomes 7/(4 − 3x).
−(7/3)ln|4 − 3x| + C
Work on either side of x = 4/3 separately.
04 · Positive denominator interval
Find ∫5/(2x + 7) dx, for x > −7/2.
Hint
Divide the numerator coefficient by 2.
Worked solution
(5/2)ln(2x + 7) + C. The argument is positive on this interval.
05 · Reciprocal square
Find ∫1/(3x + 2)² dx. Is a logarithm needed?
Hint
The exponent is −2, not −1.
Worked solution
−1/[3(3x + 2)] + C. This uses the power rule, on a valid interval; no logarithm is needed.
05 / Integrate sine and cosine with linear arguments
Retain the usual trig sign as well as 1/a.
∫sin(ax + b) dx = −cos(ax + b)/a + C
∫cos(ax + b) dx = sin(ax + b)/a + C
a ≠ 0 and angles in radians.
Find ∫3sin(1 − 2x) dx.Worked example
The sine primitive contributes a minus sign
Start with −cos(1 − 2x).
The inner slope is −2
Dividing by −2 cancels that sign.
(3/2)cos(1 − 2x) + C
Differentiate to get 3sin(1 − 2x).
06 · Cosine with a shift
Find ∫4cos(3x + 2) dx.
Hint
The shift stays inside the sine.
Worked solution
(4/3)sin(3x + 2) + C.
07 · Negative inner slope
Find ∫cos(4 − x) dx.
Hint
The inner derivative is −1.
Worked solution
−sin(4 − x) + C.
06 / Extend all the standard trig patterns
The same linear factor applies.
∫sec²(ax + b) dx = tan(ax + b)/a + C
∫cosec²(ax + b) dx = −cot(ax + b)/a + C
∫sec(ax + b)tan(ax + b) dx = sec(ax + b)/a + C
∫cosec(ax + b)cot(ax + b) dx = −cosec(ax + b)/a + C
a ≠ 0; work between poles, with radians.
08 · Secant squared
Find ∫5sec²(2x − 1) dx.
Hint
Divide by the inner slope 2.
Worked solution
(5/2)tan(2x − 1) + C, on a valid interval.
09 · A reciprocal trig product
Find ∫2cosec(3x)cot(3x) dx.
Hint
The cosecant primitive has a minus sign.
Worked solution
−(2/3)cosec(3x) + C, away from sin(3x) = 0.
07 / Evaluate exact definite integrals
Keep the factor before substituting the bounds.
Evaluate ∫₀¹ 6/(2x + 3) dx.Worked example
A primitive is 3ln(2x + 3)
The denominator is positive throughout [0,1].
3ln5 − 3ln3
Substitute upper then lower bounds.
3ln(5/3)
Combine logs only after keeping their coefficients.
10 · Exact cosine integral
Evaluate ∫₀^(π/4) 2cos(2x) dx.
Hint
A primitive is sin(2x).
Worked solution
sin(π/2) − sin0 = 1.
11 · Exact exponential integral
Evaluate ∫₀¹ 3e⁻²ˣ dx.
Hint
A primitive is −(3/2)e⁻²ˣ.
Worked solution
(3/2)(1 − e⁻²), a positive value.
08 / Find an unknown bound
First integrate, then solve the resulting equation.
Find k > 0 if ∫₀ᵏ e²ˣ dx = 4.Worked example
(e²ᵏ − 1)/2 = 4
Include the 1/2 factor before solving.
e²ᵏ = 9
Rearrange.
k = (ln9)/2 = ln3
The positive integrand makes the integral strictly increasing in k.
12 · A logarithmic bound
Find k > 0 if ∫₀ᵏ 2/(x + 1) dx = 2ln3.
Hint
The primitive is 2ln(x + 1).
Worked solution
2ln(k + 1) = 2ln3, so k + 1 = 3 and k = 2.
09 / Solve for a parameter
Do not discard valid parameter values.
Find all real a if ∫₀¹ (ax + 2)² dx = 19/3.Worked example
Expand: a²x² + 4ax + 4
This also handles a = 0 without division by a.
a²/3 + 2a + 4 = 19/3
Integrate term by term.
a² + 6a − 7 = 0
Multiply by 3 and rearrange.
(a − 1)(a + 7) = 0
Both a = 1 and a = −7 are valid; this polynomial has no domain exclusion.
13 · The zero-slope case
Find ∫e^(0x + 4) dx.
Hint
The integrand is constant.
Worked solution
e⁴x + C. Do not use the 1/a formula at a = 0.
10 / Recognise when the shortcut does not apply
A variable inner derivative is not a constant factor.
The rule here works because the inner derivative a is constant. For ∫eˣ² dx, writing eˣ²/(2x) does not undo the chain rule: differentiating 1/(2x) creates an extra term. A different approach is needed; not every integral has an elementary antiderivative.
Definite logarithmic or trig integrals must also avoid poles throughout the interval, even if the endpoint formulas are finite.
14 · Nonlinear power
Is ∫(x² + 1)³ dx equal to (x² + 1)⁴/(8x)?
Hint
The proposed coefficient 1/(8x) is variable.
Worked solution
No. Its derivative has an extra term from differentiating 1/(8x). You can instead expand the cubic and integrate the resulting polynomial.
15 · Crossing a denominator zero
Can ∫₀² 1/(2x − 1) dx be evaluated by logarithm endpoint subtraction?
Hint
Check x = 1/2.
Worked solution
No. The integrand has a pole inside the interval, so that direct definite-integral step is invalid; the ordinary improper integral diverges.
11 / Check by differentiation
The final derivative must match exactly.
- Find the inner slope a and check a ≠ 0.
- Use a standard outer antiderivative.
- Divide by a, retaining signs and coefficients.
- Separate the reciprocal case and check domains.
- Differentiate to verify, or evaluate valid bounds exactly.
16 · Repair an answer
A student gives ∫sin(5x − 2) dx = −5cos(5x − 2) + C. Repair it.
Hint
Their derivative is 25 times the required sine.
Worked solution
−(1/5)cos(5x − 2) + C. Integration divides by the inner slope instead of multiplying.
Section 1 of 11 · Undo the inner derivative