How do you integrate a rational function like $\frac{3x + 5}{x^2 + 3x + 2}$? The numerator's degree is less than the denominator's, so polynomial division isn't needed — but the integrand still doesn't match any standard form. The trick is to break the fraction into simpler pieces. Just as $3/8$ can be written as $1/2 - 1/8$, a rational function can be decomposed into a sum of simpler fractions with linear denominators. This technique, called partial fractions, transforms intimidating integrals into a collection of basic logarithmic integrals. It's the bridge that connects polynomial algebra to logarithmic integration.
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We know $\frac{3x + 5}{x^2 + 3x + 2} = \frac{3x + 5}{(x+1)(x+2)}$. If we could write this as $\frac{A}{x+1} + \frac{B}{x+2}$ for some constants $A$ and $B$, how would that make integration easier?
Core Content
Step 1 — Factor the denominator completely.
Step 2 — Set up the decomposition. For each distinct linear factor $(x - a)$, include a term $\frac{A}{x - a}$. For a repeated factor $(x - a)^2$, include $\frac{A}{x - a} + \frac{B}{(x - a)^2}$.
Step 3 — Clear fractions. Multiply both sides by the common denominator.
Step 4 — Find the constants. Either substitute convenient values of $x$ or equate coefficients of like powers.
Step 5 — Integrate each term.
Example: Decompose $\frac{3x + 5}{(x+1)(x+2)}$.
3x + 5 = A(x+2) + B(x+1)
Let x = -1: 3(-1) + 5 = A(1) → A = 2
Let x = -2: 3(-2) + 5 = B(-1) → B = 1
So: (3x+5)/((x+1)(x+2)) = 2/(x+1) + 1/(x+2)
Integrating:
∫ (3x+5)/((x+1)(x+2)) dx = 2ln|x+1| + ln|x+2| + C
When the denominator has a repeated factor like $(x - a)^2$, the decomposition requires two terms:
P(x)/(x-a)^2 = A/(x-a) + B/(x-a)^2
Example: Decompose $\frac{x + 3}{(x - 1)^2}$.
x + 3 = A(x-1) + B
Let x = 1: 1 + 3 = B → B = 4
Compare x coefficients: 1 = A → A = 1
So: (x+3)/(x-1)^2 = 1/(x-1) + 4/(x-1)^2
Integrating:
∫ = ln|x-1| - 4/(x-1) + C
Note: The second term integrates using the power rule: $\int (x-1)^{-2} \, dx = -(x-1)^{-1} + C$.
Find $\int \frac{2x + 1}{x^2 + x - 2} \, dx$.
The indefinite integral.
Step 1: Factor: x^2 + x - 2 = (x+2)(x-1)
Step 2: (2x+1)/((x+2)(x-1)) = A/(x+2) + B/(x-1)
2x + 1 = A(x-1) + B(x+2)
x = 1: 3 = 3B → B = 1
x = -2: -3 = -3A → A = 1
Step 3: ∫ = ∫ 1/(x+2) dx + ∫ 1/(x-1) dx
= ln|x+2| + ln|x-1| + C
$\ln|x+2| + \ln|x-1| + C = \ln|(x+2)(x-1)| + C$.
Find $\int \frac{5x - 1}{x^2 - x - 2} \, dx$.
Answer:
$x^2 - x - 2 = (x-2)(x+1)$. $\frac{5x-1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$.
$5x - 1 = A(x+1) + B(x-2)$. $x = 2$: $9 = 3A \Rightarrow A = 3$. $x = -1$: $-6 = -3B \Rightarrow B = 2$.
$\int = 3\ln|x-2| + 2\ln|x+1| + C$.
$\frac{P}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$
$\frac{P}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$
Substitute convenient x values
$\int \frac{A}{x-a} dx = A\ln|x-a| + C$
Writing $\frac{3x + 5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$ transforms a complex fraction into two simple ones. Each simple fraction integrates to a logarithm: $\int \frac{A}{x+1} \, dx = A\ln|x+1| + C$ and $\int \frac{B}{x+2} \, dx = B\ln|x+2| + C$. Without partial fractions, we would have no standard technique for integrating $\frac{3x+5}{x^2+3x+2}$. The decomposition is possible because any proper rational function with distinct linear factors in the denominator can be uniquely written as a sum of simpler fractions — this is the Fundamental Theorem of Partial Fractions, guaranteed by the algebra of polynomials.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Decompose $\frac{2x + 3}{x^2 - x - 6}$ into partial fractions and hence find $\int \frac{2x + 3}{x^2 - x - 6} \, dx$. Show all working. 3 MARKS
9. Decompose $\frac{3x + 1}{(x - 1)^2}$ into partial fractions and integrate. Show all working. 4 MARKS
10. In pharmacokinetics, the concentration of a drug in the bloodstream can be modelled by a rational function. Explain why partial fractions are essential for solving such models, and describe what each term in the partial fraction decomposition represents physically (e.g., absorption vs elimination). 3 MARKS
1. $\frac{4}{(x-2)(x+2)} = \frac{1}{x-2} - \frac{1}{x+2}$. $\int = \ln|x-2| - \ln|x+2| + C = \ln|\frac{x-2}{x+2}| + C$.
2. $x^2 - 2x + 1 = (x-1)^2$. $\frac{x+2}{(x-1)^2} = \frac{1}{x-1} + \frac{3}{(x-1)^2}$. $\int = \ln|x-1| - \frac{3}{x-1} + C$.
3. $x^2 + 5x + 6 = (x+2)(x+3)$. $\frac{3x+7}{(x+2)(x+3)} = \frac{1}{x+2} + \frac{2}{x+3}$. $\int = \ln|x+2| + 2\ln|x+3| + C$.
1. Two terms are needed because a single term $\frac{A}{x-a}$ cannot produce the $x^1$ term in the numerator when cleared.
2. Perform polynomial long division first to get a polynomial plus a proper rational function.
3. $\frac{1}{x^2-1} = \frac{1}{2}(\frac{1}{x-1} - \frac{1}{x+1})$. $\int = \frac{1}{2}(\ln|x-1| - \ln|x+1|) + C = \frac{1}{2}\ln|\frac{x-1}{x+1}| + C$.
Q8 (3 marks): $x^2 - x - 6 = (x-3)(x+2)$ [0.5]. $\frac{2x+3}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}$, so $2x+3 = A(x+2) + B(x-3)$ [0.5]. $x = 3$: $9 = 5A \Rightarrow A = 9/5$; $x = -2$: $-1 = -5B \Rightarrow B = 1/5$ [1]. $\int = \frac{9}{5}\ln|x-3| + \frac{1}{5}\ln|x+2| + C$ [1].
Q9 (4 marks): $\frac{3x+1}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$ [1]. $3x+1 = A(x-1) + B$ [0.5]. $x = 1$: $4 = B$ [0.5]. Compare $x$ coefficients: $3 = A$ [0.5]. So $\frac{3}{x-1} + \frac{4}{(x-1)^2}$ [0.5]. $\int = 3\ln|x-1| - \frac{4}{x-1} + C$ [1].
Q10 (3 marks): Drug concentration models often involve rational functions from solving linear differential equations with multiple exponential processes [1]. Partial fractions decompose these into terms like $\frac{A}{s+k_1}$ and $\frac{B}{s+k_2}$ (in Laplace domain) or $\frac{A}{t+\tau_1}$ in time domain [1]. Each term represents a distinct kinetic process: one for drug absorption from the gut, another for elimination by the kidneys, possibly a third for tissue distribution [1].
Climb platforms by decomposing rational functions and integrating the results. Pool: lesson 10.
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