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Mastering Integration: Algebra, Surds, and Index Laws

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    Vu Hung
    Twitter

Problem Statement

In HSC Mathematics, the most common trap students fall into is trying to integrate a complex expression immediately. In reality, integration is often the final, easiest step of a problem. The true challenge lies in the prerequisite algebraic manipulation: transforming an un-integrable expression into a standard form using index laws, surd expansion, or algebraic division.

Without a solid grasp of these algebraic fundamentals, many integrals in Extension 1 and Extension 2 remain completely inaccessible.

Consider the following integral, which requires careful algebraic manipulation and the application of index laws before any calculus can occur:

Evaluate the indefinite integral: I=∫(x2−x)2x3 dxI = \int \frac{(x^2 - \sqrt{x})^2}{x^3} \, dx

(a) Expand the numerator using algebraic techniques. (b) Use index laws to divide each term in the numerator by the denominator, converting all surds into fractional indices. (c) Evaluate the integral, expressing your final answer with positive fractional indices and surds where appropriate.

Hints

  • Part (a): Use the perfect square expansion formula: (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2. Remember that (x)2=x(\sqrt{x})^2 = x.
  • Part (b): Separate the single fraction into three separate fractions. Use the index law xmxn=xm−n\frac{x^m}{x^n} = x^{m-n}. Keep in mind that x=x1/2\sqrt{x} = x^{1/2}.
  • Part (c): Now that the expression is a sum of individual power terms, apply the standard power rule for integration: ∫xn dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C. Be careful when adding 11 to negative and fractional powers.

Solutions

Part (a): Expanding the Numerator

  1. The numerator is (x2−x)2(x^2 - \sqrt{x})^2.
  2. Let a=x2a = x^2 and b=xb = \sqrt{x}. Apply (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2: (x2)2−2(x2)(x)+(x)2(x^2)^2 - 2(x^2)(\sqrt{x}) + (\sqrt{x})^2
  3. Simplify using index laws (noting x=x1/2\sqrt{x} = x^{1/2}):
    • (x2)2=x4(x^2)^2 = x^4
    • 2(x2)(x1/2)=2x2+1/2=2x5/22(x^2)(x^{1/2}) = 2x^{2 + 1/2} = 2x^{5/2}
    • (x)2=x(\sqrt{x})^2 = x
  4. The expanded numerator is: x4−2x5/2+xx^4 - 2x^{5/2} + x.

Part (b): Applying Index Laws

  1. Substitute the expanded numerator back into the integral expression: x4−2x5/2+xx3\frac{x^4 - 2x^{5/2} + x}{x^3}
  2. Split the fraction into three individual terms: x4x3−2x5/2x3+xx3\frac{x^4}{x^3} - \frac{2x^{5/2}}{x^3} + \frac{x}{x^3}
  3. Apply the division index law (xm/xn=xm−nx^m / x^n = x^{m-n}):
    • First term: x4−3=x1=xx^{4 - 3} = x^1 = x
    • Second term: 2x5/2−3=2x5/2−6/2=2x−1/22x^{5/2 - 3} = 2x^{5/2 - 6/2} = 2x^{-1/2}
    • Third term: x1−3=x−2x^{1 - 3} = x^{-2}
  4. The prepared expression is: x−2x−1/2+x−2x - 2x^{-1/2} + x^{-2}.

Part (c): Evaluating the Integral

  1. Now, integrate the prepared expression term by term: I=∫(x−2x−1/2+x−2) dxI = \int \left( x - 2x^{-1/2} + x^{-2} \right) \, dx
  2. Apply the power rule ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}:
    • ∫x dx=x22\int x \, dx = \frac{x^2}{2}
    • ∫−2x−1/2 dx=−2(x1/21/2)=−4x1/2\int -2x^{-1/2} \, dx = -2 \left( \frac{x^{1/2}}{1/2} \right) = -4x^{1/2}
    • ∫x−2 dx=x−1−1=−x−1\int x^{-2} \, dx = \frac{x^{-1}}{-1} = -x^{-1}
  3. Combine the terms and add the constant of integration CC: I=12x2−4x1/2−x−1+CI = \frac{1}{2}x^2 - 4x^{1/2} - x^{-1} + C
  4. Convert fractional and negative indices back into surd and fraction notation for the final answer: I=x22−4x−1x+CI = \frac{x^2}{2} - 4\sqrt{x} - \frac{1}{x} + C

Takeaways

  • Preparation is Key: Never attempt to integrate a product or quotient of polynomial/surd terms directly. The power rule ∫xndx\int x^n dx only applies to individual terms of the form axnax^n.
  • Index Laws are Mandatory: Converting surds (xn=x1/n\sqrt[n]{x} = x^{1/n}) and denominators (1xn=x−n\frac{1}{x^n} = x^{-n}) into standard index form is a non-negotiable step in calculus.
  • Patience Prevents Errors: By breaking the problem down into expansion, index simplification, and finally integration, you minimise the risk of careless arithmetic errors.

Further Readings

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