Mathematics Advanced • Year 12 • Module 6 • Lesson 2

Integrating Power Functions

Build fluency integrating fractional and negative powers by rewriting roots, reciprocals and brackets in power form first.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Rewrite each in the form xⁿ:

√x = x^______ ³√x = x^______ 1/x² = x^______ 1/√x = x^______

Q1.2 The power rule fails for n = ______; in that case ∫ (1/x) dx = ____________ + C.

Q1.3 Before integrating (x² + 1)/x, divide each term by x to obtain ____________________, which is a sum of powers ready for the power rule.

Stuck? Revisit lesson § Formula Reference and § Expanding Before Integrating.

2. Worked example — ∫ (x³ + 2x)/x dx

Follow every line. Each step has a reason on the right.

Problem. Find ∫ (x³ + 2x)/x dx.

Step 1 — Divide each term in the numerator by the denominator.

(x³ + 2x)/x = x³/x + 2x/x = x² + 2

Reason: the power rule needs each term as a pure power of x.

Step 2 — Integrate term by term.

∫(x² + 2) dx = x³/3 + 2x + C

Reason: ∫x² dx = x³/3 and ∫2 dx = 2x.

Step 3 — Check by differentiating.

d/dx(x³/3 + 2x + C) = x² + 2 = (x³ + 2x)/x ✓

Answer. ∫ (x³ + 2x)/x dx = x³/3 + 2x + C.

3. Faded example — fill in the missing steps

Find ∫ (√x + 1/√x)² dx. Fill in each blank line. 4 marks

Step 1 — Expand the bracket.

(√x + 1/√x)² = (√x)² + 2(√x)(1/√x) + (1/√x)²

= x + ______ + ______

Step 2 — Rewrite each term as a power of x.

x = x¹ 2 = 2x⁰ 1/x = x^______

Step 3 — Integrate term by term.

∫ x dx = ____________ ∫ 2 dx = ____________ ∫ x⁻¹ dx = ____________

Step 4 — Combine with a single + C.

∫(√x + 1/√x)² dx = __________________________________ + C

Stuck? The middle term simplifies because √x · 1/√x = 1, so 2(√x)(1/√x) = 2.

4. Graduated practice — integrate each function

For each integral, first rewrite as a sum of powers of x, then apply the power rule. Always include + C.

Foundation — direct rewriting (4 questions)

Q	Integral	Answer (with + C)
4.1 1	∫ x⁻³ dx
4.2 1	∫ √x dx
4.3 1	∫ x^3/2 dx
4.4 1	∫ (1/√x) dx

Standard — typical HSC difficulty (6 questions)

Show the rewriting step on the line below each part.

4.5 ∫ ³√(x²) dx 2 marks

4.6 ∫ (x⁴ − 2x)/x² dx 2 marks

4.7 ∫ x(x + 3) dx 2 marks

4.8 ∫ (x + 1)² dx 2 marks

4.9 ∫ (√x − 4/x²) dx 2 marks

4.10 ∫ (x³ + 4x² − x)/x² dx 2 marks

Extension — combine concepts (2 questions)

4.11 A tank fills at rate dV/dt = 4/√t L/min. Find V(t) given V(1) = 0, then use V(t) to find the volume in the tank at t = 4. 3 marks

4.12 A student writes "∫ x⁻¹ dx = x⁰/0 + C". State why this is invalid in one line, give the correct antiderivative, and explain (one sentence) why the absolute value is needed. 3 marks

Stuck on 4.12? The "exception" is the case n = −1; covered in the Formula Reference panel.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (∫ x⁻³ dx) I added 1 to the exponent before dividing, getting x⁻²/(−2) + C = −1/(2x²) + C.

For 4.2 (∫ √x dx) I wrote √x as x^1/2 first, then used (1/2) + 1 = 3/2 in the power rule.

For 4.4 (∫ 1/√x dx) my answer 2√x + C checks: d/dx(2√x) = 2 · (1/(2√x)) = 1/√x ✓.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Rewriting

√x = x^1/2; ³√x = x^1/3; 1/x² = x⁻²; 1/√x = x^−1/2.

Q1.2 — The exception

Power rule fails for n = −1. ∫(1/x) dx = ln|x| + C.

Q1.3 — Pre-processing

(x² + 1)/x = x + 1/x, ready for the power rule (with the 1/x case handled by ln|x|).

Q3 — Faded example ∫(√x + 1/√x)² dx

Step 1: (√x + 1/√x)² = x + 2 + 1/x.
Step 2: 1/x = x⁻¹.
Step 3: ∫x dx = x²/2; ∫2 dx = 2x; ∫x⁻¹ dx = ln|x|.
Step 4: x²/2 + 2x + ln|x| + C.

Q4.1 — ∫ x⁻³ dx

x⁻²/(−2) + C = −1/(2x²) + C.

Q4.2 — ∫ √x dx

∫x^1/2 dx = x^3/2/(3/2) + C = (2/3) x^3/2 + C.

Q4.3 — ∫ x^3/2 dx

x^5/2/(5/2) + C = (2/5) x^5/2 + C.

Q4.4 — ∫ (1/√x) dx

∫x^−1/2 dx = x^1/2/(1/2) + C = 2√x + C.

Q4.5 — ∫ ³√(x²) dx

∫x^2/3 dx = x^5/3/(5/3) + C = (3/5) x^5/3 + C.

Q4.6 — ∫ (x⁴ − 2x)/x² dx

Rewrite: (x⁴ − 2x)/x² = x² − 2x⁻¹. Integrate: x³/3 − 2 ln|x| + C.

Q4.7 — ∫ x(x + 3) dx

= ∫(x² + 3x) dx = x³/3 + 3x²/2 + C.

Q4.8 — ∫ (x + 1)² dx

(x+1)² = x² + 2x + 1, so ∫(x² + 2x + 1) dx = x³/3 + x² + x + C. (Equivalent form (x+1)³/3 + C.)

Q4.9 — ∫ (√x − 4/x²) dx

= ∫ (x^1/2 − 4 x⁻²) dx = (2/3) x^3/2 − 4 · x⁻¹/(−1) + C = (2/3) x^3/2 + 4/x + C.

Q4.10 — ∫ (x³ + 4x² − x)/x² dx

Rewrite: x + 4 − x⁻¹. ∫(x + 4 − 1/x) dx = x²/2 + 4x − ln|x| + C.

Q4.11 — Tank with dV/dt = 4/√t, V(1) = 0; V(4)

V(t) = ∫ 4 t^−1/2 dt = 4 · 2 t^1/2 + C = 8√t + C. V(1) = 0 ⇒ 0 = 8 + C ⇒ C = −8. V(t) = 8√t − 8. V(4) = 16 − 8 = 8 L.

Q4.12 — Invalid use of power rule on x⁻¹

The power rule requires n ≠ −1; here n = −1 makes the denominator (n + 1) zero, so the rule cannot be applied.
Correct antiderivative: ∫ x⁻¹ dx = ln|x| + C.
The absolute value lets the antiderivative cover both x > 0 and x < 0, matching the natural domain of 1/x (which excludes only x = 0).