01 · Find the constant match
For ∫5x(x² + 4)² dx, what is k when g = x² + 4?
Hint
Compare 5x with 2x.
Worked solution
k = 5/2, giving (5/6)(x² + 4)³ + C.
Understand · explore · practise
Reverse chain rule integration
Use the reverse chain rule to integrate composite powers and exponentials. Identify the inner derivative, match constant factors, check domains and practise definite integrals.
Before you startThe chain rule, standard integrals and integration of linear arguments.
01 / Look for the inner derivative
The extra factor can make a composite integral simple.
∫k g′(x)[g(x)]ⁿ dx = k[g(x)]ⁿ⁺¹/(n + 1) + C
k is constant; n ≠ −1. Work on a valid real interval.
Differentiating a power of g creates g′ as a factor. Run that idea backwards: spot the inner function, differentiate it, and compare the remaining factor with g′. A constant multiple can be corrected outside the integral.
Find the inner derivativeExplore
At x = 1: g = 4; g′ = 2; supplied factor = 4.
Integrand value = 64.
A primitive is (2/3)g³ + C.
Its derivative value is 64, matching the integrand.
The factor match is an algebraic identity. At a point where g′ is zero, both sides still agree; no numerical division by g′ is required. These inner functions stay positive at the model’s inputs.
02 / Match the factor before integrating
The ratio must be constant as an expression.
Find ∫7x(x² + 3)⁴ dx.Worked example
Choose g = x² + 3, so g′ = 2x
The inner derivative is already present up to a constant.
7x = (7/2)×2x
The match is k = 7/2.
(7/2)∫g′g⁴ dx = (7/2)g⁵/5
Integrate the outer power.
(7/10)(x² + 3)⁵ + C
Differentiate: (7/10)×5×2x = 7x.
Watch: match the factor and reverse the chain rule
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Include negative powers
The new exponent may be negative too.
Find ∫6x/(x² + 5)³ dx.Worked example
Rewrite as 6x(x² + 5)⁻³
Use n = −3.
g = x² + 5 and 6x = 3g′
The constant match is 3.
3g⁻²/(−2) + C
The new exponent is −2.
−3/[2(x² + 5)²] + C
The denominator is never zero for real x.
02 · Reciprocal square
Find ∫4x/(x² + 2)² dx.
Hint
k = 2 and n = −2.
Worked solution
−2/(x² + 2) + C.
03 · A cubic inner function
Find ∫9x²/(x³ + 1)² dx.
Hint
g′ = 3x².
Worked solution
−3/(x³ + 1) + C, on either interval avoiding x = −1.
04 / Handle square roots as powers
Keep the outer exponent and inner derivative separate.
Find ∫x√(x² + 1) dx.Worked example
g = x² + 1 and x = (1/2)g′
The inner function is positive for every real x.
n = 1/2, so n + 1 = 3/2
Divide by the new exponent.
(1/2)×(2/3)g³ᐟ² + C
Combine the constant factors.
(1/3)(x² + 1)³ᐟ² + C
Differentiate to check.
04 · A reciprocal square root
Find ∫3x/√(x² + 4) dx.
Hint
The outer exponent is −1/2.
Worked solution
3√(x² + 4) + C.
05 · A restricted radical domain
Find ∫x√(x² − 4) dx for x > 2.
Hint
Use g = x² − 4.
Worked solution
(1/3)(x² − 4)³ᐟ² + C. The stated interval keeps the square root real and smooth.
05 / Reverse an exponential chain
The outer exponential integrates to itself.
∫k g′(x)eᵍ⁽ˣ⁾ dx = k eᵍ⁽ˣ⁾ + C
Find ∫5x eˣ²⁺³ dx.Worked example
g = x² + 3; g′ = 2x
The exponential has the inner function g.
5x = (5/2)g′
Only a constant adjustment is needed.
(5/2)eˣ²⁺³ + C
Differentiate to recover 5x eˣ²⁺³.
06 · Trig inside an exponential
Find ∫3cos x eˢⁱⁿˣ dx.
Hint
Take g = sin x.
Worked solution
3eˢⁱⁿˣ + C.
07 · Exponential inside a power
Find ∫5eˣ(2 + eˣ)³ dx.
Hint
Take g = 2 + eˣ.
Worked solution
(5/4)(2 + eˣ)⁴ + C.
06 / Use a trigonometric inner function
Differentiate the whole proposed inner function.
Find ∫4cos x(2 + sin x)³ dx.Worked example
g = 2 + sin x; g′ = cos x
The shift differentiates to zero.
The supplied factor is 4g′
Use k = 4 and n = 3.
(2 + sin x)⁴ + C
The factors 4 and 1/4 cancel.
Find ∫9sec³x tan x dx.Worked example
g = sec x; g′ = sec x tan x
One sec x factor belongs to g′.
9sec³x tan x = 9g′g²
The remaining outer power is 2, not 3.
3sec³x + C
Check: 3×3sec²x×sec x tan x = 9sec³x tan x.
08 · Negative derivative sign
Find ∫6sin x(3 + cos x)² dx.
Hint
The inner derivative is −sin x.
Worked solution
−2(3 + cos x)³ + C.
09 · Count the remaining power
In ∫sec⁵x tan x dx with g = sec x, what is the remaining power of g?
Hint
Remove one sec x for g′.
Worked solution
g⁴ remains, so the integral is sec⁵x/5 + C on a valid interval.
07 / Evaluate without expanding
Once a primitive is known, use the original x bounds.
Evaluate ∫₀² x(x² + 1)² dx.Worked example
A primitive is (x² + 1)³/6
Use g′ = 2x and the outer exponent 2.
At x = 2, the numerator is 5³ = 125
Keep the original upper x bound.
At x = 0, the numerator is 1
Subtract the lower value.
(125 − 1)/6 = 62/3
The result is exact.
10 · A definite exponential chain
Evaluate ∫₀^(π/2) cos x eˢⁱⁿˣ dx.
Hint
A primitive is eˢⁱⁿˣ.
Worked solution
e − 1.
11 · A symmetric cancellation
Evaluate ∫₋₁¹ x(x² + 3)² dx.
Hint
The primitive (x² + 3)³/6 has equal endpoint values.
Worked solution
0. This is a signed integral; the nonzero regions on either side cancel.
08 / Solve for an unknown bound
Use the domain or sign restriction when solving.
Find k ≥ 0 if ∫₀ᵏ 2x eˣ² dx = e⁹ − 1.Worked example
A primitive is eˣ²
The factor 2x is the exact inner derivative.
eᵏ² − 1 = e⁹ − 1
Evaluate at k and zero.
k² = 9
The exponential is one-to-one.
k = 3
The restriction k ≥ 0 selects the root.
12 · A zero derivative at one point
Does g′(0) = 0 invalidate the reverse-chain primitive for g = x² + 3?
Hint
The chain-rule identity is still true at zero.
Worked solution
No. The algebraic factor match holds for every x, including zero. It does not require dividing numerical values by g′(0).
09 / Separate the logarithmic exception
When the outer power is −1, use a different formula.
∫k g′(x)/g(x) dx = k ln|g(x)| + C
Work on a connected interval where g is nonzero.
The power formula would divide by n + 1 = 0. The logarithmic form has the correct derivative instead; the next lesson studies this pattern and its domain checks in detail.
13 · Recognise the exception
Find ∫4x/(x² + 3) dx.
Hint
The inner function is positive and k = 2.
Worked solution
2ln(x² + 3) + C.
10 / Do not invent a constant match
A variable ratio needs another method.
For ∫(x² + 3)² dx, the factor 2x is missing. Dividing a proposed primitive by 2x introduces a variable coefficient and an extra derivative term. You can expand this particular polynomial instead.
Being able to divide two values at one input does not establish a constant ratio of functions. Check the expression, not a single point.
14 · Detect the missing factor
Does ∫x²(x² + 3)⁴ dx fit the simple reverse-power rule with g = x² + 3?
Hint
Compare x² with 2x as functions.
Worked solution
No constant multiple of 2x equals x² for every x. This direct shortcut does not apply; the polynomial could be expanded.
15 · An alternative route
Integrate (x² + 3)² by expanding.
Hint
The expansion is x⁴ + 6x² + 9.
Worked solution
x⁵/5 + 2x³ + 9x + C.
11 / Spot, match, integrate, verify
Differentiation checks both the outer power and the inner factor.
- Choose an inner function g.
- Calculate g′ exactly, including its sign.
- Check the remaining factor is a constant multiple of g′.
- Integrate the outer power or exponential.
- Separate n = −1 and respect domains.
- Differentiate the whole answer to check.
16 · Repair a coefficient
A student gives ∫8x(x² + 1)³ dx = 2(x² + 1)⁴ + C. Repair the answer.
Hint
Differentiate their proposed coefficient.
Worked solution
The correct result is (x² + 1)⁴ + C, whose derivative is 4(x² + 1)³×2x. The proposed result is twice as large.
Section 1 of 11 · Look for the inner derivative