Mathematics Advanced • Year 12 • Module 6 • Lesson 3
Integrating Exponentials and Logarithms
Build fluency in integrating ex, ekx, ax and 1/x — the rules that fill the gaps left by the power rule.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the three rules from this lesson:
∫ ex dx = __________________ + C
∫ ekx dx = __________________ + C (k ≠ 0)
∫ (1/x) dx = __________________ + C
Q1.2 Why does the antiderivative of 1/x use the absolute value inside the logarithm? (One sentence.) __________________________________________________________________
Q1.3 Complete: ∫ ax dx = ____________ + C, where a > 0 and a ≠ 1.
2. Worked example — ∫ (2 e3x + 4/x) dx
Follow every line. Each step has a reason on the right.
Problem. Find ∫ (2 e3x + 4/x) dx.
Step 1 — Sum rule, term by term.
∫(2 e^{3x} + 4/x) dx = ∫2 e^{3x} dx + ∫(4/x) dx
Reason: the integral of a sum is the sum of the integrals.
Step 2 — Apply each rule separately.
∫2 e^{3x} dx = 2 · (1/3) e^{3x} = (2/3) e^{3x}
∫(4/x) dx = 4 ln|x|
Reason: ∫e^{kx} dx = (1/k) e^{kx}; ∫(1/x) dx = ln|x|.
Step 3 — Combine with a single + C.
(2/3) e^{3x} + 4 ln|x| + C
Step 4 — Check.
d/dx[(2/3) e^{3x} + 4 ln|x| + C] = (2/3)·3 e^{3x} + 4 · (1/x) = 2 e^{3x} + 4/x ✓
Answer. (2/3) e3x + 4 ln|x| + C.
3. Faded example — fill in the missing steps
Find ∫ (e−x + 3/x + x²) dx. Fill in each blank line. 4 marks
Step 1 — Sum rule, term by term:
∫e^{−x} dx uses ∫e^{kx} dx = (1/k) e^{kx} with k = −1:
∫e^{−x} dx = ____________
∫(3/x) dx = ____________
∫x² dx = ____________
Step 2 — Combine with a single + C:
∫(e−x + 3/x + x²) dx = ______________________________________ + C
Step 3 — Check by differentiating. You should recover the original integrand:
d/dx(your answer) = ______________________________________
4. Graduated practice — integrate each function
For each integral, identify which rule applies and write the antiderivative. Always include + C.
Foundation — single-rule (4 questions)
| Q | Integral | Answer (with + C) |
|---|---|---|
| 4.1 1 | ∫ e4x dx | |
| 4.2 1 | ∫ 5 e−2x dx | |
| 4.3 1 | ∫ (3/x) dx | |
| 4.4 1 | ∫ 2x dx |
Standard — typical HSC difficulty (6 questions)
Show working on the line below each part.
4.5 ∫ (ex + 1/x + x³) dx 2 marks
4.6 ∫ (3 e2x − 5/x + 4x) dx 2 marks
4.7 ∫ (e0.5x + 6/x) dx 2 marks
4.8 ∫ (e−x + x−1 + x1/2) dx 2 marks
4.9 ∫ (4 ex − 7/x) dx 2 marks
4.10 ∫ (e3x + e−3x) dx 2 marks
Extension — combine concepts (2 questions)
4.11 A drug decays in the body at rate dA/dt = −0.5 e−0.5t mg/h. Find A(t) given A(0) = 2 mg. 3 marks
4.12 Show that d/dx (3x/ln 3) = 3x, and hence write down ∫ 3x dx. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Three rules
∫ ex dx = ex + C; ∫ ekx dx = (1/k) ekx + C; ∫ (1/x) dx = ln|x| + C.
Q1.2 — Why |x|
1/x is defined for all x ≠ 0, but ln x is only defined for x > 0; using ln|x| extends the antiderivative to both positive and negative x.
Q1.3 — General base
∫ ax dx = ax/ln a + C.
Q3 — Faded example ∫(e−x + 3/x + x²) dx
Step 1: ∫e−x dx = −e−x; ∫(3/x) dx = 3 ln|x|; ∫x² dx = x³/3.
Step 2: −e−x + 3 ln|x| + x³/3 + C.
Step 3 check: d/dx(−e−x + 3 ln|x| + x³/3) = e−x + 3/x + x² ✓.
Q4.1 — ∫ e4x dx
(1/4) e4x + C.
Q4.2 — ∫ 5 e−2x dx
5 · (1/−2) e−2x + C = −(5/2) e−2x + C.
Q4.3 — ∫ (3/x) dx
3 ln|x| + C.
Q4.4 — ∫ 2x dx
2x/ln 2 + C.
Q4.5 — ∫ (ex + 1/x + x³) dx
ex + ln|x| + x⁴/4 + C.
Q4.6 — ∫ (3 e2x − 5/x + 4x) dx
(3/2) e2x − 5 ln|x| + 2x² + C.
Q4.7 — ∫ (e0.5x + 6/x) dx
(1/0.5) e0.5x + 6 ln|x| + C = 2 e0.5x + 6 ln|x| + C.
Q4.8 — ∫ (e−x + x−1 + x1/2) dx
−e−x + ln|x| + (2/3) x3/2 + C.
Q4.9 — ∫ (4 ex − 7/x) dx
4 ex − 7 ln|x| + C.
Q4.10 — ∫ (e3x + e−3x) dx
(1/3) e3x + (1/−3) e−3x + C = (1/3) e3x − (1/3) e−3x + C.
Q4.11 — Drug decay, A(0) = 2
A(t) = ∫(−0.5 e−0.5t) dt = (−0.5) · (1/−0.5) e−0.5t + C = e−0.5t + C. A(0) = 2: 2 = 1 + C ⇒ C = 1. A(t) = e−0.5t + 1.
Q4.12 — Differentiate 3x/ln 3
d/dx (3x/ln 3) = (1/ln 3) · 3x · ln 3 = 3x ✓. Hence ∫ 3x dx = 3x/ln 3 + C.