NCEA Level 3Integrating rational functionsAS 91579

Integrating fractions: when it is ln, and when it is not

What this lesson covers

  • ln f(x) is the answer only when the numerator is the derivative of the denominator.
  • Other fractions are split, divided, or integrated by reverse chain rule.
Key idea

Not every fraction is ln

  • The 2025 error: ln of the bottom, every time.
  • That works for one kind of fraction.
  • Tell the kinds apart first.
Worked example

The ln case: ∫ 2x/(x² + 5) dx

  • Differentiate the bottom
  • It matches the top
  1. ∫ 2x/(x² + 5) dx
  2. d/dx (x² + 5) = 2xderivative of bottom
  3. top = 2x: a match
  4. = ln(x² + 5) + cx² + 5 > 0
  5. d/dx ln(x² + 5) = 2x/(x² + 5)
Worked example

A constant multiple still counts

  • Top = multiple of bottom′
  • Adjust the number in front
  1. ∫ x/(x² + 5) dx
  2. x = (1/2)(2x)half the derivative
  3. = (1/2) ∫ 2x/(x² + 5) dx
  4. = (1/2)ln(x² + 5) + c
  5. ∫ 3/(2x + 1) dx
  6. d/dx (2x + 1) = 2derivative of bottom
  7. = (3/2) ∫ 2/(2x + 1) dx3 = (3/2) × 2
  8. = (3/2)ln|2x + 1| + cbottom can be < 0
Key idea

Four routes for a fraction

  • First: differentiate the bottom.
  • Match: ln. One-term bottom: split.
  • Top power as high: divide. Bracket: chain.
FractionTestRoute
2x/(x² + 1)top = bottom′ln
(x² + 1)/xsingle-term bottomsplit
(x² + 2x + 3)/(x + 1)top power ≥ bottomdivide
1/(2x + 1)³a bracket powerreverse chain
Worked example

Split: ∫ (x² + 3x − 2)/x dx

  • Divide each term by x
  • Integrate term by term
  1. ∫ (x² + 3x − 2)/x dx
  2. = ∫ (x + 3 − 2/x) dxdivide each term by x
  3. ∫ x dx = x²/2
  4. ∫ 3 dx = 3x
  5. ∫ −2/x dx = −2ln|x|the only log
  6. = x²/2 + 3x − 2ln|x| + c
Worked example

Split: ∫ (4x + 1)/x² dx

  • 4/x gives a log
  • 1/x² is x⁻², a power
  1. ∫ (4x + 1)/x² dx
  2. = ∫ (4/x + x⁻²) dxdivide each term by x²
  3. ∫ 4/x dx = 4ln|x|a log
  4. ∫ x⁻² dx = x⁻¹/(−1) = −1/xa power
  5. = 4ln|x| − 1/x + c
Worked example

Divide: ∫ (x² + 2x + 3)/(x + 1) dx

  • Top power ≥ bottom
  • Check the division
  1. (x² + 2x + 3) ÷ (x + 1)
  2. x(x + 1) = x² + xx into x²
  3. (x² + 2x + 3) − (x² + x) = x + 3
  4. 1(x + 1) = x + 11 into x
  5. (x + 3) − (x + 1) = 2
  6. quotient x + 1, remainder 2
  7. (x + 1)² + 2 = x² + 2x + 3
  8. ∫ (x + 1 + 2/(x + 1)) dx
  9. = x²/2 + x + 2ln|x + 1| + c
Worked example

Reverse chain: ∫ 6/(2x + 1)³ dx

  • Write it as a power
  • Divide by power and by 2
  1. ∫ 6/(2x + 1)³ dx
  2. = ∫ 6(2x + 1)⁻³ dxas a power
  3. = 6(2x + 1)⁻²/(−2 × 2) + c
  4. = −3/(2(2x + 1)²) + c6/(−4) = −3/2
  5. d/dx gives 6(2x + 1)⁻³check
Worked example

The "everything is ln" error

  • Split, divide or chain
  • Differentiate to test it
  1. ∫ 1/x² dx = ln(x²)differentiates to 2/x
  2. ∫ x⁻² dx = −1/x + ca power
  3. ∫ (x² + 1)/x dx = ln xsplit first
  4. ∫ (x + 1/x) dx = x²/2 + ln|x| + c
  5. ∫ x/(x + 2) dx = ln|x + 2|divide first
  6. ∫ (1 − 2/(x + 2)) dxx = (x + 2) − 2
  7. = x − 2ln|x + 2| + c
  8. ∫ (2x + 1)⁻³ dxreverse chain, not ln
  9. = −1/(4(2x + 1)²) + c
Worked example

Solving for an unknown limit

  • From 0 to k, equals ln 5
  • k > 0: reject −2
  1. ∫ 2x/(x² + 1) dx = ln(x² + 1)
  2. ln(k² + 1) − ln 1 = ln 5limits k and 0
  3. ln(k² + 1) = ln 5ln 1 = 0
  4. k² + 1 = 5equal logs
  5. k² = 4, so k = ±2
  6. k = −2k > 0
  7. k = 2
Key idea

Where the marks are

  • Achieved: rearrange, spot the ln case.
  • Merit: keep the algebra right afterwards.
  • Excellence: long division first.
Summary

What to take away

  • Differentiate the bottom first, then choose the route.
Top = bottom′ln of the bottom
One-term bottomsplit
Top power ≥divide first
Bracket powerreverse chain rule
Checkdifferentiate back
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Decide the route first.
  • One answer has no log at all.
IntegrateFind
16x/(x² + 4)its integral
2(x³ − 5)/x²its integral
3(x² + 3x + 5)/(x + 2)its integral
48/(4x − 1)²its integral
Show answer

1. 3ln(x² + 4) + c. 2. x²/2 + 5/x + c, with no ln at all. 3. x + 1 remainder 3, so x²/2 + x + 3ln|x + 2| + c. 4. −2/(4x − 1) + c.

Worked example

The answers: questions 1 and 2

  1. 1. bottom′ = 2x, top 6x = 3 × 2xa multiple: ln
  2. = 3ln(x² + 4) + c
  3. 2. ∫ (x − 5x⁻²) dxdivide each term by x²
  4. = x²/2 − 5x⁻¹/(−1) + cpower rule
  5. = x²/2 + 5/x + cno log
Worked example

The answers: questions 3 and 4

  1. 3. (x² + 3x + 5) ÷ (x + 2)
  2. x(x + 2) = x² + 2xleaves x + 5
  3. 1(x + 2) = x + 2leaves 3
  4. ∫ (x + 1 + 3/(x + 2)) dx
  5. = x²/2 + x + 3ln|x + 2| + c
  6. 4. ∫ 8(4x − 1)⁻² dxas a power
  7. = 8(4x − 1)⁻¹/(−1 × 4) + cdivide by −1 and by 4
  8. = −2/(4x − 1) + c

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