IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Integrating with Exponential & Logarithmic Functions

The natural cases ∫e^x dx = e^x + c and ∫1/x dx = ln|x| + c you’ve met before — and ∫a^x dx = a^x/ln a + c generalises to any base. The real power comes from one extra pattern: reverse chain rule for ratios. If the numerator is (a multiple of) the derivative of the denominator — the f′/f pattern — the integral is ln|f| + c. Spot that pattern early and a wide class of “ugly looking” integrals becomes trivial.

📘 What you need to know

Three formula-booklet antiderivatives: ∫e^x dx = e^x + c; ∫1/x dx = ln|x| + c; ∫a^x dx = 1ln a a^x + c.
Watch the asymmetry vs differentiation: derivative of a^x is a^x · ln a (MULTIPLY by ln a); antiderivative DIVIDES by ln a. Same ln a factor, opposite operation.
Reverse chain rule pattern: ∫ f′(x)f(x) dx = ln|f(x)| + c. This is the single most-useful shortcut in this topic — and it works for any f, not just polynomials.
Linear-argument formulas: ∫e^ax+b dx = (1/a) e^ax+b + c; ∫ 1/(ax+b) dx = (1/a) ln|ax+b| + c. The 1/a is reverse chain rule “adjust and compensate”.
Modulus on log antiderivatives: always write ln|f(x)|. Drop the modulus ONLY when you can prove f(x) is strictly positive on the integration interval.
Definite-integral cheat-codes: e⁰ = 1, ln(1) = 0, ln(e) = 1, e^{ln a} = a. Examiners pick limits like ln(3) or 1/e precisely so the algebra collapses.
Log-base-a antiderivative (not in formula booklet, derivable): ∫ 1x ln a dx = log_a|x| + c. Comes from reading d/dx[log_a x] = 1/(x ln a) backwards.
+c on every indefinite. Always. Even a perfect calculation loses a mark without it.

The base antiderivatives — and the f′/f shortcut

Three formula-booklet antiderivatives + f′/f reverse chain rule ∫e^x dx = e^x + c | ∫ 1x dx = ln|x| + c | ∫a^x dx = a^xln a + c

∫ f′(x)f(x) dx = ln|f(x)| + c

The ∫1/x dx = ln|x| + c result has a beautiful geometric meaning: the area under the curve y = 1/x between x = 1 and x = e is exactly 1.

The number e ≈ 2.718 is DEFINED so that the area under y = 1/x from 1 to e is exactly 1. Equivalently: ln(e) = 1. The green shaded region’s exact area comes from the FTC and the antiderivative ln|x|. Every natural-log calculation you do is, geometrically, an area under 1/x.

The f′/f shortcut — your most valuable trick: whenever you see a fraction where the numerator equals (or is a constant multiple of) the derivative of the denominator, the antiderivative is just ln|denominator| + c. Pattern works for polynomial denominators (2x/(x²+5)), trig denominators (cos x/(sin x+2)), exp denominators (e^x/(e^x+1)) — anywhere the algebra lines up.

Linear arguments and general bases

Linear arg — natural exp and ln

∫e^ax+b dx = 1a e^ax+b + c

∫ 1ax+b dx = 1a ln|ax+b| + c

divide by the inner coefficient — “adjust and compensate” reverse chain rule

Linear arg — general base a^px+q

∫a^px+q dx = 1p ln a a^px+q + c

two factors in the denominator: p from the chain rule, and ln a from the base change

🧭 Recipe — integrate any exp or log expression

Check for the f′/f pattern FIRST: is the numerator (a multiple of) the derivative of the denominator? If yes, the answer is ln|denominator| + c, possibly with a constant in front. This saves a lot of work — always look for it before anything else.
Identify the function type: e^x, a^x, 1/x, or 1/(linear). Note whether the argument is just x or linear (ax+b). Pull any constant numerator outside the integral.
Quote the formula-booklet antiderivative for the matching form. For general bases a^x, DIVIDE by ln a (not multiply — that’s the derivative).
For linear arguments, “adjust and compensate” by 1/a on the exp side, or by 1/a ahead of the ln on the log side. For general-base linear arguments, both the 1/p AND the 1/ln a appear.
For definite integrals, apply the bounds using exact values (e⁰ = 1, e^{ln k} = k, ln(1) = 0, ln(e) = 1). For indefinite, add “+c” and keep modulus on the log unless f(x) is provably positive on the interval.

Worked examples

WE 1

Linear-argument natural exponential

Find ∫e^3x+1 dx.

Step 1 — identify the form e^(ax+b) with a = 3, b = 1 Step 2 — quote the linear-argument formula ∫ e^(ax+b) dx = (1/a) e^(ax+b) + c Step 3 — substitute a = 3 ∫ e^(3x+1) dx = (1/3) e^(3x+1) + c ∫ e^(3x+1) dx = 13 e^(3x+1) + c verify: d/dx[(1/3) e^(3x+1)] = (1/3) · 3 · e^(3x+1) = e^(3x+1). ✓ the 3 from the chain rule cancels the 1/3. The constant +1 inside the exponent has no effect on the 1/a factor — only the coefficient of x matters.

WE 2

Definite general-base exponential with linear argument

Evaluate ∫₀¹ 5^2x dx, giving your answer in exact form.

Step 1 — identify a = 5, p = 2 (linear argument 2x) ∫ a^(px+q) dx = (1/(p ln a)) a^(px+q) + c Step 2 — antiderivative ∫ 5^(2x) dx = (1/(2 ln 5)) · 5^(2x) + c Step 3 — apply FTC bounds 0 and 1 [(1/(2 ln 5)) · 5^(2x)] from 0 to 1 = (1/(2 ln 5))(5^2 – 5^0) = (1/(2 ln 5))(25 – 1) = 24/(2 ln 5) = 12/ln 5 ∫ from 0 to 1 of 5^(2x) dx = 12ln 5 two coefficients in the denominator of the antiderivative: 2 (from chain rule on 2x) and ln 5 (from the general base). Both must appear. Keep ln 5 exact — don’t approximate to 1.609.

WE 3

Reverse chain rule f′/f — polynomial denominator

Find ∫ 2xx² + 5 dx.

Step 1 — check the f’/f pattern let f(x) = x² + 5 f'(x) = 2x ← exactly the numerator! so this is ∫ f'(x)/f(x) dx Step 2 — apply the f’/f shortcut ∫ f'(x)/f(x) dx = ln|f(x)| + c ∫ 2x/(x² + 5) dx = ln|x² + 5| + c Step 3 — can we drop the modulus? x² + 5 ≥ 5 > 0 for all real x so yes — drop the modulus safely ∫ 2x/(x² + 5) dx = ln(x² + 5) + c always check whether f(x) is strictly positive before dropping the modulus. For x² + (positive constant), it is — always. For x² – 4 it isn’t — keep the bars on. This single-step shortcut saves a substitution u = x² + 5; spot it early.

WE 4

Linear-argument natural log antiderivative

Find ∫ 53x + 2 dx.

Step 1 — pull the constant 5 outside ∫ 5/(3x + 2) dx = 5 ∫ 1/(3x + 2) dx Step 2 — apply linear-argument formula with a = 3, b = 2 ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + c ∫ 1/(3x + 2) dx = (1/3) ln|3x + 2| + c Step 3 — multiply by the outer 5 5 · (1/3) ln|3x + 2| + c = (5/3) ln|3x + 2| + c ∫ 5/(3x + 2) dx = 53 ln|3x + 2| + c KEEP the modulus bars — 3x + 2 can be negative (for x < -2/3). Same f’/f logic works here too: derivative of (3x + 2) is 3, numerator is 5 = (5/3) · 3, so the (5/3) factor and the ln antiderivative emerge from the same pattern.

WE 5

Reverse chain rule f′/f — trig denominator

Find ∫ cos xsin x + 2 dx.

Step 1 — check the f’/f pattern let f(x) = sin x + 2 f'(x) = cos x ← exactly the numerator! Step 2 — apply the f’/f shortcut ∫ cos x/(sin x + 2) dx = ln|sin x + 2| + c Step 3 — can we drop the modulus? -1 ≤ sin x ≤ 1, so 1 ≤ sin x + 2 ≤ 3 strictly positive — drop the modulus ∫ cos x/(sin x + 2) dx = ln(sin x + 2) + c the f’/f pattern is base-agnostic — it works just as well for trig as for polynomials. A student who knows only substitution might spend two minutes on u = sin x + 2, du = cos x dx; a student who SPOTS f’/f writes the answer in one line. Recognition over computation.

WE 6

Definite natural exponential — limit chosen to collapse

Evaluate ∫₀^{ln 3} e^2x dx, giving your answer in exact form.

Step 1 — antiderivative via linear-argument formula ∫ e^(ax) dx = (1/a) e^(ax) + c ∫ e^(2x) dx = (1/2) e^(2x) + c Step 2 — apply the FTC with limits 0 and ln 3 [(1/2) e^(2x)] from 0 to ln 3 = (1/2)(e^(2 ln 3) – e^0) Step 3 — simplify e^(2 ln 3) using e^(ln k) = k e^(2 ln 3) = e^(ln 3²) = e^(ln 9) = 9 (log power rule) e^0 = 1 Step 4 — combine = (1/2)(9 – 1) = (1/2)(8) = 4 ∫ from 0 to ln 3 of e^(2x) dx = 4 the upper limit ln 3 is chosen DELIBERATELY so e^(2 · ln 3) collapses to a clean integer. When you see a limit like ln(k) on an e^x integral, suspect a clean answer is coming — and use e^(ln k) = k or the power rule e^(c ln k) = k^c.

💡 Top tips

Scan for the f′/f pattern FIRST on any integral involving a fraction. It’s by far the most common reverse-chain-rule pattern in this topic and saves substitution work.
Quote the formula booklet for ∫e^x, ∫1/x, and ∫a^x. Even simple cases earn marks for citation. For general base, DIVIDE by ln a — derivative multiplies, antiderivative divides.
Use exact-value collapses in definite integrals: e^{ln k} = k, e⁰ = 1, ln(1) = 0, ln(e^k) = k. Examiners chose those limits for a reason.
Modulus discipline: write ln|f(x)| always, then drop bars only after checking f(x) > 0 on the integration interval. For x² + (positive), it’s safe; for (3x+2) or (sin x − 5), it’s not always.
Check by differentiating — for these antiderivatives, the chain-rule coefficients are easy to miss. If d/dx[your answer] reproduces the integrand, you’re right.

⚠ Common mistakes

Multiplying by ln a instead of dividing: ∫a^x dx = a^x/ln a + c, NOT a^x · ln a + c. The “multiply” version is the DERIVATIVE.
Dropping the modulus from ln without checking positivity: ∫1/(3x+2) dx = (1/3) ln|3x+2| + c, NOT (1/3) ln(3x+2) + c in general.
Missing the 1/a on linear arguments: ∫e^3x+1 dx = (1/3) e^3x+1 + c, NOT e^3x+1 + c. Same for ln antiderivatives.
Failing to spot f′/f and resorting to substitution unnecessarily. ∫(2x)/(x²+5) dx is one line if you see the pattern; three lines via u = x²+5.
Forgetting +c — costs a mark every single time, even when the rest is flawless.

Up next: Integration by Substitution. Reverse chain rule worked when you could SPOT the f′/f pattern by inspection — but for trickier integrands where the pattern is hidden, substitution makes it explicit. You’ll set u = (inner function), find du/dx, change variables (including the limits for definite integrals), then integrate in u. The substitution is sometimes given in the question, sometimes you have to spot it yourself.

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