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Partial Fractions and Complex Roots

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    Vu Hung
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Problem Statement

The method of partial fractions is an essential integration technique for breaking down rational functions. Usually, you factorise the denominator into linear factors like (x−a)(x−b)(x-a)(x-b). However, what happens when the denominator contains an irreducible quadratic, like x2+4x^2 + 4?

In standard real numbers, it cannot be factorised. You would traditionally use the form Ax+Bx2+4\frac{Ax + B}{x^2 + 4}. But in HSC Mathematics Extension 2, we have access to the complex number ii. We can factorise x2+4x^2 + 4 over the complex field into (x−2i)(x+2i)(x - 2i)(x + 2i) and use standard linear partial fraction techniques!

Consider the integral: I=∫8x(x2+4) dxI = \int \frac{8}{x(x^2 + 4)} \, dx

(a) Factorise the denominator completely over the complex field.

(b) Using the complex number ii, decompose the integrand into partial fractions of the form: Ax+Bx−2i+Cx+2i\frac{A}{x} + \frac{B}{x - 2i} + \frac{C}{x + 2i} Find the complex constants A,BA, B, and CC.

(c) Evaluate the integral. (Note: While you can integrate the complex fractions to get complex logarithms, for HSC purposes, recombine the conjugate pairs back into a real fraction before integrating to yield a real arctan⁡\arctan or ln⁡\ln function).

Hints

  • Part (a): Use the difference of two squares technique, treating x2+4x^2 + 4 as x2−(−4)x^2 - (-4) or x2−(2i)2x^2 - (2i)^2.
  • Part (b): Multiply both sides by the common denominator x(x−2i)(x+2i)x(x - 2i)(x + 2i) to get the basic equation: 8=A(x−2i)(x+2i)+Bx(x+2i)+Cx(x−2i)8 = A(x - 2i)(x + 2i) + Bx(x + 2i) + Cx(x - 2i). Use the "cover-up" method: substitute the roots x=0x=0, x=2ix=2i, and x=−2ix=-2i to find A,B,A, B, and CC.
  • Part (c): Once you have the fractions, take the two terms involving ii and combine them back over a common real denominator (x2+4x^2 + 4). You will find that the imaginary parts cancel out nicely. Then integrate using standard real calculus (the natural log and arctan⁡\arctan rules).

Solutions

Part (a): Complex Factorisation

  1. The denominator is x(x2+4)x(x^2 + 4).
  2. We can rewrite x2+4x^2 + 4 as a difference of two squares using i2=−1i^2 = -1: x2+4=x2−(−4)=x2−(2i)2x^2 + 4 = x^2 - (-4) = x^2 - (2i)^2
  3. Factorise: x2−(2i)2=(x−2i)(x+2i)x^2 - (2i)^2 = (x - 2i)(x + 2i)
  4. The completely factorised denominator is x(x−2i)(x+2i)x(x - 2i)(x + 2i).

Part (b): Partial Fraction Decomposition

  1. Set up the identity: 8x(x−2i)(x+2i)=Ax+Bx−2i+Cx+2i\frac{8}{x(x - 2i)(x + 2i)} = \frac{A}{x} + \frac{B}{x - 2i} + \frac{C}{x + 2i}
  2. Multiply by the common denominator: 8=A(x−2i)(x+2i)+Bx(x+2i)+Cx(x−2i)8 = A(x - 2i)(x + 2i) + Bx(x + 2i) + Cx(x - 2i)
  3. Find AA by letting x=0x = 0: 8=A(−2i)(2i)+0+08 = A(-2i)(2i) + 0 + 0 8=A(−4i2)=A(−4(−1))=4A8 = A(-4i^2) = A(-4(-1)) = 4A A=2A = 2
  4. Find BB by letting x=2ix = 2i: 8=0+B(2i)(2i+2i)+08 = 0 + B(2i)(2i + 2i) + 0 8=B(2i)(4i)=B(8i2)=−8B8 = B(2i)(4i) = B(8i^2) = -8B B=−1B = -1
  5. Find CC by letting x=−2ix = -2i: 8=0+0+C(−2i)(−2i−2i)8 = 0 + 0 + C(-2i)(-2i - 2i) 8=C(−2i)(−4i)=C(8i2)=−8C8 = C(-2i)(-4i) = C(8i^2) = -8C C=−1C = -1
  6. The decomposition is: 2x−1x−2i−1x+2i\frac{2}{x} - \frac{1}{x - 2i} - \frac{1}{x + 2i}

Part (c): Evaluating the Integral

  1. Recombine the complex conjugate pair to return to real numbers before integrating: −1x−2i−1x+2i=−(x+2i+x−2i(x−2i)(x+2i))-\frac{1}{x - 2i} - \frac{1}{x + 2i} = -\left( \frac{x + 2i + x - 2i}{(x - 2i)(x + 2i)} \right) =−(2xx2+4)= -\left( \frac{2x}{x^2 + 4} \right)
  2. Our integrand is now: ∫(2x−2xx2+4) dx\int \left( \frac{2}{x} - \frac{2x}{x^2 + 4} \right) \, dx
  3. Notice how the Ax+BAx+B form automatically emerged! The numerator 2x2x is exactly the derivative of the denominator x2+4x^2 + 4.
  4. Integrate using the rule ∫f′(x)f(x)dx=ln⁡∣f(x)∣\int \frac{f'(x)}{f(x)} dx = \ln|f(x)|:
    • ∫2x dx=2ln⁡∣x∣\int \frac{2}{x} \, dx = 2\ln|x|
    • ∫2xx2+4 dx=ln⁡(x2+4)\int \frac{2x}{x^2 + 4} \, dx = \ln(x^2 + 4) (no absolute value needed as x2+4>0x^2+4>0)
  5. Combine the results: I=2ln⁡∣x∣−ln⁡(x2+4)+CI = 2\ln|x| - \ln(x^2 + 4) + C
  6. Use logarithm laws to simplify: I=ln⁡(x2)−ln⁡(x2+4)+CI = \ln(x^2) - \ln(x^2 + 4) + C I=ln⁡(x2x2+4)+CI = \ln \left( \frac{x^2}{x^2 + 4} \right) + C

Takeaways

  • The Cover-Up Method on Steroids: The standard "cover-up" method for linear partial fractions works perfectly with complex roots. This is often faster and less error-prone than solving simultaneous equations for A,B,CA, B, C using the Ax+Bx2+4\frac{Ax+B}{x^2+4} real-number method.
  • Conjugate Pairs: When you decompose a real fraction using complex roots, the resulting complex constants (BB and CC) will always form conjugate pairs. When recombined, all imaginary terms cancel out, leaving a pure real fraction.
  • Multiple Paths to the Solution: Integration is an art. Choosing to route your algebra through the complex plane to simplify a real-world integral is a hallmark of an advanced mathematician.

Further Readings

Connect with me

For more advanced tricks bridging complex numbers and calculus, head to Vu's Maths Hub and check out the Extension 2 booklets. Join me as I solve these exact problems on my YouTube channel. Don't forget to connect on LinkedIn or read my latest syllabus deep-dives on Substack.