VCE Specialist Mathematics (Year 12) · Calculus · 20m read

Integration Techniques — Substitution and Partial Fractions

Ever stared at an integral and thought, "There's no way this is in my formula sheet"? That's where substitution and partial fractions come in — they're your toolkit for breaking down tricky integrals into pieces you can actually handle. Substitution swaps variables to simplify messy expressions, while partial fractions splits complicated fractions into simpler chunks. These techniques unlock integrals you'll see constantly in calculus applications and VCE exams, turning impossible-looking problems into straightforward ones.

Learning objective: Apply substitution and partial fractions to evaluate integrals not in standard form.

By the end of this lesson you can:

• I apply substitution $u = g(x)$ to transform integrals to standard form.
• I decompose rational functions into partial fractions and integrate.
• I evaluate definite integrals using these techniques.

Integration techniques are a toolbox. Substitution reverses the chain rule; partial fractions break a big fraction into easy pieces.

Integration by substitution

### The substitution method

If an integral has the form $\int f(g(x)) \cdot g'(x)\,dx$, let $u = g(x)$:

$$\frac{du}{dx} = g'(x) \implies du = g'(x)\,dx$$

The integral becomes $\int f(u)\,du$.

Step-by-step:
1. Identify $u = g(x)$ (usually the "inner" function).
2. Find $du/dx$ and rearrange for $dx$.
3. Substitute: replace $g(x)$ with $u$ and $g'(x)dx$ with $du$.
4. Integrate with respect to $u$.
5. Back-substitute: replace $u$ with $g(x)$.

For definite integrals, also change the limits when substituting.

### Partial fractions

For rational integrals $\int \frac{P(x)}{Q(x)}\,dx$ where degree of $P <$ degree of $Q$:

Case 1 — distinct linear factors: $\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$

Case 2 — repeated factor: $\frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$

Case 3 — irreducible quadratic: $\frac{P(x)}{(x^2+1)} = \frac{Ax+B}{x^2+1}$ → leads to $\arctan$ integrals.

After decomposition, integrate each term using $\int \frac{1}{x-a}\,dx = \ln|x-a| + C$.

Quick reference

| Technique | When |
|---|---|
| Substitution u = g(x) | Integrand contains g′(x) |
| Partial fractions | Rational functions |
| Trig identities | Powers of sin/cos |
| Change limits | Definite integrals with substitution |

Key ideas at a glance

• Look for f(g(x))g′(x): Substitution.
• Factor denominator: Partial fractions.
• Change limits: With u.
• Check by differentiating: Verify.

Worked example (Medium)

Evaluate $\int \frac{2x}{x^2 + 1}\,dx$.

1. Identify substitution. The numerator $2x$ is the derivative of $x^2 + 1$. u = x^2 + 1, \quad du = 2x\,dx
2. Substitute. \int \frac{2x}{x^2+1}\,dx = \int \frac{1}{u}\,du = \ln|u| + C
3. Back-substitute. = \ln|x^2 + 1| + C = \ln(x^2+1) + C \quad (\text{since } x^2+1 > 0)

Answer: $\ln(x^2+1) + C$.

Worked example (Medium)

Evaluate $\displaystyle\int x \sqrt{x^2 + 4}\, dx$ by substitution.

1. Substitution. Let $u = x^2 + 4$. du = 2x\, dx,; x\, dx = du/2
2. Rewrite integral. In terms of $u$. \int \sqrt{u} \cdot du/2 = \tfrac{1}{2} \int u^{1/2}\, du
3. Integrate. Power rule. \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} = \tfrac{1}{3} u^{3/2}
4. Back-substitute. Replace $u$. \tfrac{1}{3}(x^2 + 4)^{3/2} + C

Answer: $\tfrac{1}{3}(x^2 + 4)^{3/2} + C$.

Worked example (Hard)

Evaluate $\int \frac{3x+1}{(x-1)(x+2)}\,dx$.

1. Partial fractions decomposition. \frac{3x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}
2. Find A and B. Multiply both sides by $(x-1)(x+2)$: 3x+1 = A(x+2) + B(x-1)
3. Use strategic substitution. $x=1$: $4 = 3A \Rightarrow A = 4/3$. $x=-2$: $-5 = -3B \Rightarrow B = 5/3$. A = \tfrac{4}{3}, \quad B = \tfrac{5}{3}
4. Integrate. \int \frac{3x+1}{(x-1)(x+2)}\,dx = \frac{4}{3}\ln|x-1| + \frac{5}{3}\ln|x+2| + C

Answer: $\frac{4}{3}\ln|x-1| + \frac{5}{3}\ln|x+2| + C$.

Common mistakes

• Not changing limits in definite integrals with substitution When you substitute $u = g(x)$, the limits are $x$-values. After substitution, the limits become $u$-values: $u = g(a)$ and $u = g(b)$. Fix: For $\int_a^b f(g(x))g'(x)\,dx$: when $x = a$, $u = g(a)$; when $x = b$, $u = g(b)$. Use these as the new limits.

Partial fractions and the algebra of rational functions

The partial fraction decomposition is not just a trick — it reflects a deep algebraic fact. Over the real numbers, every polynomial factors into linear and irreducible quadratic factors (by the Fundamental Theorem of Algebra applied to the reals). The partial fraction theorem states that any proper rational function $P(x)/Q(x)$ can be uniquely decomposed into a sum of simpler fractions whose denominators are powers of these irreducible factors.

This is analogous to writing integers as sums of unit fractions in number theory, and it connects to the residue calculus in complex analysis. There, the coefficients $A$, $B$ in the decomposition correspond to residues of the function at its poles, and integration of rational functions around closed curves is governed by the Residue Theorem.

In linear algebra, partial fractions underpin the Heaviside cover-up method and are essential for computing inverse Laplace transforms — the primary tool for solving linear ODEs in engineering. Given $F(s) = \frac{P(s)}{Q(s)}$, decomposition into partial fractions allows term-by-term inversion using standard Laplace tables, converting frequency-domain transfer functions back to time-domain solutions.

At university, you will also encounter partial fractions over the complex numbers, where every factor is linear and the decomposition simplifies further.

Practice questions

1. Using the substitution $u = 3x^2 - 1$, evaluate $\int x(3x^2 - 1)^4\,dx$.

   (2 marks)

2. Decompose $\frac{5}{(x+1)(x-4)}$ into partial fractions and integrate.

   (4 marks)

3. Decompose $\dfrac{3x + 1}{(x - 1)(x + 2)}$ into partial fractions.

   (3 marks)

Summary

• Substitution: $u = g(x)$, $du = g'(x)\,dx$. Transform → integrate → back-substitute.
• Definite integrals: change limits to $u$-values when substituting.
• Partial fractions: $\frac{P}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$.
• Find constants by substituting strategic $x$ values.
• Integrate using $\int \frac{1}{x-a}\,dx = \ln|x-a| + C$.