01 · Positive denominator
Find ∫8x/(x² + 7) dx.
Hint
Compare 8x with 2x.
Worked solution
4ln(x² + 7) + C. The denominator is positive for all real x.
Understand · explore · practise
Logarithmic integrals
Integrate f prime over f using ln absolute value. Learn logarithmic integrals of rational and trigonometric functions, check domain intervals and practise exact definite integrals.
Before you startDifferentiating logarithms, the chain rule and standard integrals.
01 / Spot the logarithmic pattern
The numerator is a constant multiple of the denominator’s derivative.
∫k g′(x)/g(x) dx = k ln|g(x)| + C
k is constant. Work on an interval where g is differentiable and nonzero.
The logarithm is the missing case of the reverse power rule: the outer exponent −1 cannot use division by n + 1. Differentiate ln|g| on either a positive or negative branch of g and the result is g′/g.
Match the numerator; check the domainExplore
At x = 3: g = 5; g′ = 6; numerator = 18.
Integrand = 3.6; derivative of the primitive = 3.6.
Primitive value without C = 4.828314.
This point is in x > 2, where g is positive.
Different constants of integration may be used on separate intervals. The graph marks excluded denominator zeros, not zeros of the numerator.
02 / Match a numerator exactly
A missing numerical factor changes the whole answer.
Find ∫6x/(x² − 4) dx.Worked example
g = x² − 4; g′ = 2x
Use the denominator as the inner function.
6x = 3g′
The constant match is 3.
3ln|x² − 4| + C
Valid separately on x < −2, −2 < x < 2 and x > 2.
Derivative = 3×2x/(x² − 4)
Check both the coefficient and denominator.
Watch: one formula, three domain intervals
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Keep the absolute value when needed
A logarithm of a negative number is not a real answer.
Find ∫3/(3x − 5) dx for x < 5/3.Worked example
The denominator is negative on this interval
ln(3x − 5) is not real here.
ln|3x − 5| = ln(5 − 3x)
Both expressions describe the same real function on the stated interval.
Derivative of ln(5 − 3x) = −3/(5 − 3x)
This equals 3/(3x − 5).
Absolute value makes the logarithm valid for nonzero negative inputs. It does not make ln 0 defined, and does not permit a definite integral to cross a pole.
02 · Find the excluded points
For ∫2x/(x² − 9) dx, list the intervals on which one primitive can be used.
Hint
Solve x² − 9 = 0.
Worked solution
x < −3, −3 < x < 3, and x > 3. A primitive on each is ln|x² − 9| + C; the constants can differ.
03 · Remove absolute value locally
Rewrite ln|x² − 9| for −3 < x < 3.
Hint
Here x² − 9 is negative.
Worked solution
ln(9 − x²).
04 / Allow a negative inner derivative
The coefficient must include its sign.
Find ∫5/(7 − 2x) dx.Worked example
g = 7 − 2x; g′ = −2
The numerator 5 is (−5/2)g′.
−(5/2)ln|7 − 2x| + C
Work on either side of x = 7/2.
Derivative = (−5/2)(−2)/(7 − 2x)
The two negative factors give the required numerator.
04 · Linear denominator
Find ∫7/(4x + 1) dx.
Hint
The denominator derivative is 4.
Worked solution
(7/4)ln|4x + 1| + C, away from x = −1/4.
05 · Repair the sign
A student writes ∫1/(2 − x) dx = ln|2 − x| + C. Correct it.
Hint
Differentiate their answer.
Worked solution
−ln|2 − x| + C. The inner derivative is −1.
05 / Use exponential and logarithmic inner functions
The same derivative test works beyond polynomials.
Find ∫4eˣ/(3 + eˣ) dx.Worked example
g = 3 + eˣ; g′ = eˣ
The numerator is 4g′.
4ln(3 + eˣ) + C
The inner function is always positive.
Find ∫1/(x ln x) dx.Worked example
g = ln x; g′ = 1/x
The numerator is the factor 1/x.
ln|ln x| + C
Real domain: 0 < x < 1 or x > 1.
06 · Exponential numerator
Find ∫6e²ˣ/(5 + e²ˣ) dx.
Hint
The inner derivative is 2e²ˣ.
Worked solution
3ln(5 + e²ˣ) + C.
07 · Logarithm inside a logarithm
Evaluate ∫ₑ^(e²) 1/(x ln x) dx.
Hint
Both endpoints lie in x > 1.
Worked solution
ln 2 − ln 1 = ln 2.
06 / Integrate tan and cot by rewriting
Use sine and cosine to reveal the pattern.
Find ∫tan x dx.Worked example
tan x = sin x/cos x
Take g = cos x.
g′ = −sin x
The numerator is −g′.
−ln|cos x| + C
Equivalent to ln|sec x| + C on intervals where cos x ≠ 0.
∫cot x dx = ln|sin x| + C
For cot x = cos x/sin x, the numerator is exactly the denominator’s derivative.
08 · A linear trig argument
Find ∫tan(3x) dx.
Hint
Include the factor from differentiating cos(3x).
Worked solution
−(1/3)ln|cos(3x)| + C, on intervals avoiding cos(3x) = 0.
09 · Shifted sine denominator
Find ∫2cos x/(4 + sin x) dx.
Hint
Take g = 4 + sin x.
Worked solution
2ln(4 + sin x) + C. This denominator is always positive.
07 / Recognise the sec and cosec logarithms
A useful multiplication creates a derivative numerator.
Derive ∫sec x dx.Worked example
Multiply by (sec x + tan x)/(sec x + tan x)
This factor is 1 wherever sec x is defined; sec x + tan x cannot be zero there.
(sec²x + sec x tan x)/(sec x + tan x)
The numerator is the derivative of the denominator.
ln|sec x + tan x| + C
Differentiate to verify.
∫cosec x dx = −ln|cosec x + cot x| + C
The derivative of cosec x + cot x is −cosec x(cosec x + cot x). Use intervals avoiding sin x = 0.
10 · A scaled secant integral
Find ∫sec(2x) dx.
Hint
Differentiate the standard logarithmic expression with argument 2x.
Worked solution
(1/2)ln|sec(2x) + tan(2x)| + C, where cos(2x) ≠ 0.
11 · Check the cosecant sign
What is the derivative of ln|cosec x + cot x|?
Hint
Factor the numerator after using the chain rule.
Worked solution
−cosec x. Its negative is therefore a primitive of cosec x.
08 / Evaluate exact logarithmic differences
Check the whole interval before applying the bounds.
Evaluate ∫₀² 6x/(x² + 5) dx.Worked example
A primitive is 3ln(x² + 5)
There are no denominator zeros.
3ln 9 − 3ln 5
Substitute upper then lower bounds.
3ln(9/5)
Combine with the quotient law; this is exact.
Evaluate ∫₀^(π/3) tan x dx.Worked example
Use −ln(cos x)
Cosine is positive throughout this interval.
−ln(1/2) + ln 1 = ln 2
Keep the exact trig value.
12 · Negative denominator interval
Evaluate ∫₀¹ 2x/(x² − 4) dx.
Hint
Use ln|x² − 4|; no zero lies in [0,1].
Worked solution
ln 3 − ln 4 = ln(3/4), a negative signed integral.
09 / Do not cross an excluded point
Endpoint substitution alone can hide a divergent integral.
The expression ln|x| has the same value at −1 and 1. That does not make ∫₋₁¹ 1/x dx equal to zero as an ordinary definite integral: 1/x is undefined at zero and its one-sided integrals diverge. This is not the same as a continuous odd function’s symmetric cancellation.
In A-level definite-integral questions, inspect the denominator on the entire interval. Do not silently cancel opposite infinities.
13 · A forbidden interval
Can you evaluate ∫₀³ 2x/(x² − 4) dx by subtracting endpoint logarithms?
Hint
Locate the denominator zero inside the interval.
Worked solution
No. x = 2 lies inside [0,3]; the ordinary improper integral does not converge.
10 / Check the pattern before taking a logarithm
A denominator by itself is not enough.
Find ∫(2x + 3)/(x² + 3x + 8) dx.Worked example
g = x² + 3x + 8; g′ = 2x + 3
Here the numerator matches exactly.
g = (x + 3/2)² + 23/4 > 0
The logarithm has a positive argument for all real x.
ln(x² + 3x + 8) + C
A derivative check confirms the match.
For ∫1/(x² + 3) dx, ln(x² + 3) is not a primitive: its derivative has a numerator 2x. You need a different method, not an invented variable multiplier.
14 · Constant or variable match?
Does ∫x²/(x² + 4) dx directly fit k g′/g with g = x² + 4 and constant k?
Hint
Compare x² with 2x as functions.
Worked solution
No. Their ratio is not constant. Algebraic division is a useful first step instead.
15 · A cubic denominator
Find ∫12x²/(x³ + 7) dx.
Hint
The denominator derivative is 3x².
Worked solution
4ln|x³ + 7| + C, separately on intervals excluding x = −∛7.
11 / Match, restrict, integrate, check
Use the numerator and domain together.
- Differentiate the proposed inner function.
- Match a constant multiple of that derivative.
- Use ln|g|, unless positivity is established.
- Keep denominator zeros outside the working interval.
- Differentiate the final answer to check its factor and sign.
16 · Find an unknown endpoint
Find b > 0 if ∫₀ᵇ 2x/(x² + 1) dx = ln 10.
Hint
The integral is ln(b² + 1).
Worked solution
b² + 1 = 10, so b = 3 using b > 0.
Section 1 of 11 · Spot the logarithmic pattern