🔬 Grade 12 · Lesson 5 of 13
The Partial-Fraction Decomposer: Splitting to Integrate
One awkward rational function is really a sum of simple ones hiding in disguise — unmask them and the integral falls out.
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The big idea: Adding fractions merges several simple pieces over a common denominator into one complicated fraction. Partial-fraction decomposition runs that process in reverse: it breaks a rational function back into a sum of simpler fractions, one per factor of the denominator. This matters for integration, because each simple piece integrates to a logarithm you already know — while the combined fraction looks intractable.
Running fraction addition backwards
You know that 1⁄(x−1) − 1⁄(x+1), placed over a common denominator, collapses into a single fraction with denominator (x−1)(x+1) = x²−1. Partial-fraction decomposition is the reverse trip: start from the messy combined fraction and recover the simple pieces.
Why bother? Because you cannot integrate 1⁄(x²−1) at a glance, but you can integrate each of 1⁄(x−1) and 1⁄(x+1) instantly — each gives a logarithm.
One fraction per factor
Factor the denominator into distinct linear pieces. Each factor (x − r) contributes one term of the form (a constant) ÷ (x − r). The whole rational function equals the sum of these terms — once we find the right constants.
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The decomposition template
For a proper fraction (numerator degree less than denominator degree) with distinct linear factors, write one term per factor with an unknown constant on top, then solve for the constants.
\[ \frac{px + q}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b} \]
Each distinct linear factor of the denominator gets its own simple fraction with a constant numerator.
🎮 Partial-Fraction Decomposer LIVE
Split one complicated fraction into simple ones you can integrate.
The cover-up method
To find the constants, clear denominators to get a polynomial identity, then substitute the value of x that makes one factor zero. Each such substitution kills all but one unknown, revealing it immediately. This shortcut is often called the cover-up method.
📝 Worked example: Decompose 1/((x − 1)(x + 1)) and hence integrate it.
- Write \( \dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1} \).
- Clear denominators: \( 1 = A(x+1) + B(x-1) \).
- Let x = 1: \( 1 = A(2) \Rightarrow A = \tfrac12 \). Let x = −1: \( 1 = B(-2) \Rightarrow B = -\tfrac12 \).
- So the fraction equals \( \tfrac12\cdot\dfrac{1}{x-1} - \tfrac12\cdot\dfrac{1}{x+1} \), and integrating each term gives \( \tfrac12\ln|x-1| - \tfrac12\ln|x+1| + C \).
✓ The integral is <strong>½ ln|(x−1)/(x+1)| + C</strong>.
\[ \int \frac{dx}{x^2 - 1} = \frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right| + C \]
Impossible to see directly, routine once decomposed: each simple piece integrates to a logarithm.
📝 Worked example: Decompose (5x − 4)/(x² − x − 2) into partial fractions.
- Factor the denominator: \( x^2 - x - 2 = (x - 2)(x + 1) \).
- Set \( \dfrac{5x - 4}{(x-2)(x+1)} = \dfrac{A}{x-2} + \dfrac{B}{x+1} \), so \( 5x - 4 = A(x+1) + B(x-2) \).
- Let x = 2: \( 10 - 4 = A(3) \Rightarrow A = 2 \). Let x = −1: \( -5 - 4 = B(-3) \Rightarrow B = 3 \).
- Check: \( 2(x+1) + 3(x-2) = 5x - 4 \). ✓
✓ The decomposition is <strong>2/(x−2) + 3/(x+1)</strong>.
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The fraction must be proper first
Partial fractions apply only when the numerator’s degree is less than the denominator’s. If it is equal or greater, do polynomial long division first, then decompose the leftover remainder fraction. Also note: repeated factors like (x−2)² and irreducible quadratics need extended templates — but distinct linear factors are the core case.
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Linear factor, meet logarithm
Every simple piece a⁄(x−r) integrates to a ln|x−r|. So decomposing a rational function into linear pieces turns a hard integral into a tidy sum of logarithms — that is the entire payoff of the technique.
Check your understanding
1. The correct partial-fraction setup for 1/((x − 1)(x + 1)) is:
Each distinct linear factor gets one term with a constant numerator: A/(x−1) + B/(x+1).
2. For 3/((x − 2)(x + 1)) = A/(x − 2) + B/(x + 1), the value of A is:
Cover-up at x = 2: A = 3/((2)+1) = 3/3 = 1.
3. What is ∫ 1/(x − 3) dx?
The integral of 1/(x − a) is ln|x − a| + C — this is why partial fractions produce logarithms.
4. Before decomposing (x³ + x)/(x² − 1), what must you do first?
The numerator's degree (3) exceeds the denominator's (2), so it's improper — do long division first, then decompose the remainder.
5. In 5/((x − 1)(x + 4)) = A/(x − 1) + B/(x + 4), the value of B is:
Cover-up at x = −4: B = 5/((−4) − 1) = 5/(−5) = −1.
✅ Key takeaways
- Partial-fraction decomposition reverses fraction addition, splitting one rational function into simpler pieces.
- For distinct linear factors, write one term A/(x − r) per factor and solve for the constants.
- The cover-up method: clear denominators, then substitute each root to isolate one constant at a time.
- Each simple piece a/(x − r) integrates to a ln|x − r|, turning a hard integral into a sum of logarithms.
- The fraction must be proper (numerator degree < denominator degree); otherwise do long division first.
- Example: 1/(x² − 1) = ½/(x − 1) − ½/(x + 1), so its integral is ½·ln|(x − 1)/(x + 1)| + C.