Cambridge 9709 · Advanced Log & Exponential Integration · Maclaurin Series
This module extends your A-Level logarithm and exponential work from P2 into the full Cambridge 9709 Pure 3 syllabus. You will master integration techniques that produce natural logarithm answers, build confidence with differential equations, and explore Maclaurin series expansions.
Recall from P2 that ∫(1/x) dx = ln|x| + c. The chain rule extension gives us:
The factor (1/a) appears because differentiating ln|ax+b| gives a/(ax+b), so we must divide by a to compensate.
When the integrand is a rational function, decompose into partial fractions first. Each term of the form A/(ax+b) integrates to (A/a) ln|ax+b|.
The derivative of ln|f(x)| is f'(x)/f(x) by the chain rule. Reversing this gives:
This is called the reverse chain rule for logarithms. To use it, the numerator must be exactly (or a constant multiple of) the derivative of the denominator.
The most important trig application is ∫tan x dx:
As with ln, the factor (1/a) compensates for the chain rule: d/dx[e^(ax+b)] = a·e^(ax+b).
Cambridge P3 requires exact answers. Leave your answer in terms of e, not as a decimal.
This follows because d/dx(aˣ) = aˣ ln a, so dividing by ln a reverses the differentiation.
A separable differential equation has the form dy/dx = f(x)·g(y). Rearrange to collect y-terms on one side, x-terms on the other, then integrate both sides.
k > 0: exponential growth. k < 0: exponential decay. N₀ is the initial value when t = 0.
The Maclaurin series is a Taylor series centred at x = 0:
Since all derivatives of eˣ equal eˣ, and e⁰ = 1, every coefficient f⁽ⁿ⁾(0)/n! = 1/n!.
Range of validity is crucial — the series diverges outside this interval.
Find ∫ 3/(4x−1) dx.
Find ∫ (x+7)/((x+1)(x+3)) dx.
Find ∫ (6x²−4)/(2x³−4x+7) dx.
Show that ∫ tan x dx = −ln|cos x| + c = ln|sec x| + c.
Evaluate ∫₀^(ln2) (e^(2x) + 1) dx, giving an exact answer.
A bacteria culture satisfies dN/dt = 0.05N. Initially N = 200. Find N when t = 20.
Find the first four terms of the Maclaurin series for e^(3x). State the range of validity.
Find the first three non-zero terms of (1+x) ln(1+x) for small x.
The chain rule requires division by the coefficient of x.
Always write ln|...| unless the domain guarantees positivity.
Always check the coefficient. The numerator must be exactly the derivative (or a multiple of it).
sin x = −d/dx(cos x), so the negative sign appears.
One arbitrary constant suffices; adding on both sides is redundant and confusing.
The series converges at x=1 (harmonic series with alternating signs) but diverges at x=−1.
Unless asked to derive from first principles, substitute into standard series and adjust.
| Expression | Integral / Result | Notes |
|---|---|---|
| ∫ 1/(ax+b) dx | (1/a) ln|ax+b| + c | Divide by coefficient a |
| ∫ 1/x dx | ln|x| + c | Special case a=1, b=0 |
| ∫ f'(x)/f(x) dx | ln|f(x)| + c | Reverse chain rule |
| ∫ tan x dx | −ln|cos x| + c = ln|sec x| + c | Standard result |
| ∫ cot x dx | ln|sin x| + c | cos x = d/dx(sin x) |
| ∫ eˣ dx | eˣ + c | Unique self-integral |
| ∫ e^(ax+b) dx | (1/a) e^(ax+b) + c | Divide by a |
| ∫ aˣ dx | aˣ / ln a + c | a > 0, a ≠ 1 |
| d/dx[ln|f(x)|] | f'(x)/f(x) | Chain rule |
| d/dx[e^(f(x))] | f'(x) e^(f(x)) | Chain rule |
| Growth/decay DE | dN/dt = kN → N = N₀ e^(kt) | k>0 growth, k<0 decay |
| eˣ Maclaurin | 1 + x + x²/2! + x³/3! + ... | Valid all x ∈ ℝ |
| ln(1+x) Maclaurin | x − x²/2 + x³/3 − x⁴/4 + ... | Valid −1 < x ≤ 1 |
| ln(1−x) Maclaurin | −x − x²/2 − x³/3 − ... | Valid −1 ≤ x < 1 |
| e^(ax) Maclaurin | 1 + ax + (ax)²/2! + (ax)³/3! + ... | Substitute ax into eˣ |
| Partial fractions form | A/(ax+b) + B/(cx+d) | For distinct linear factors |
Claim: d/dx[ln|x|] = 1/x for all x ≠ 0.
Case 1: x > 0. Then |x| = x, so d/dx[ln x] = 1/x by the standard derivative. ✓
Case 2: x < 0. Then |x| = −x, so ln|x| = ln(−x). By the chain rule:
d/dx[ln(−x)] = 1/(−x) · (−1) = 1/x ✓
Therefore d/dx[ln|x|] = 1/x for all x ≠ 0. Reversing the differentiation gives ∫ 1/x dx = ln|x| + c. □
Extension: Replacing x with (ax+b): d/dx[ln|ax+b|] = a/(ax+b), so ∫ 1/(ax+b) dx = (1/a)ln|ax+b| + c.
Setup: Let f(x) = eˣ. We seek f(x) = Σ aₙxⁿ where aₙ = f⁽ⁿ⁾(0)/n!.
Key observation: For all n ≥ 0, f⁽ⁿ⁾(x) = eˣ (eˣ is its own derivative of all orders).
Therefore f⁽ⁿ⁾(0) = e⁰ = 1 for all n.
eˣ = Σ (xⁿ/n!) = 1 + x + x²/2! + x³/3! + x⁴/4! + ...
Convergence: The ratio test shows |aₙ₊₁/aₙ| = |x|/(n+1) → 0 as n → ∞ for any fixed x. Therefore the series converges absolutely for all x ∈ ℝ. □
Method 1 — Direct differentiation: Let f(x) = ln(1+x).
f(0) = 0. f'(x) = 1/(1+x) → f'(0) = 1.
f''(x) = −1/(1+x)² → f''(0) = −1. So a₂ = −1/2.
f'''(x) = 2/(1+x)³ → f'''(0) = 2. So a₃ = 2/6 = 1/3.
f⁽ⁿ⁾(x) = (−1)^(n+1) · (n−1)! / (1+x)ⁿ → f⁽ⁿ⁾(0) = (−1)^(n+1)(n−1)! (n ≥ 1).
So aₙ = f⁽ⁿ⁾(0)/n! = (−1)^(n+1)/n.
ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + ... = Σ (−1)^(n+1) xⁿ/n, n≥1
Method 2 — Integrating the geometric series:
1/(1+x) = 1 − x + x² − x³ + ... (geometric series, |x| < 1).
Integrating term by term: ln(1+x) = x − x²/2 + x³/3 − ... (adding constant 0 since ln1=0).
Validity: Converges for −1 < x ≤ 1 (conditional convergence at x=1, divergence at x=−1). □
Enter answers as decimals (±0.001) or exact coefficients. For ln answers, enter the coefficient only where indicated.
Find ∫ (3x+11)/((x+1)(x+5)) dx, giving your answer in the form a ln|x+1| + b ln|x+5| + c. State the values of a and b.
Evaluate ∫₁³ x/(x²+3) dx, giving your answer in exact form.
Solve the differential equation dy/dx = (y+1)/(2x+1), given that y = 2 when x = 0. Give your answer in the form y = f(x).
Find the first three non-zero terms of the Maclaurin series for e^(2x) ln(1+x). State the range of values of x for which the expansion is valid.
The mass M grams of a radioactive substance satisfies dM/dt = −0.03M. Initially M = 500 g. Find (a) the mass after 10 years, (b) the time at which M = 100 g.
Show that ∫₀^(ln3) e^(2x)/(e^(2x)+1) dx = (1/2) ln(5/2).
Use the series for ln(1+x) to show that, for small x, ln((1+2x)/(1−2x)) ≈ 4x + (32/3)x³. State the range of x for which this approximation holds.
A curve satisfies the differential equation dy/dx = y²(2x+1) and passes through (0, 1). Find y in terms of x, and find the value of x where y = 4.
Given that ∫₀^a 8/(2x+1)(2x+3) dx = ln 3, find the value of a.
By first finding the partial fraction decomposition, find ∫ (5x+3)/((x+1)²(x+2)) dx.
A particle moves so that its velocity v at time t satisfies dv/dt = v cos t. Initially v = 2. Find v in terms of t, and find the value of v when t = π/2.
Find the first three non-zero terms in the Maclaurin expansion of e^x / (1+x). Hence find an approximation for ∫₀^0.1 eˣ/(1+x) dx.
The variables x and y satisfy the differential equation (1+x²) dy/dx = xy, and y = 2 when x = 0. (i) Solve the equation, giving y² in terms of x. (ii) Find the exact value of y when x = 1.