How to Solve: Integration (Basic, By Substitution, By Parts, Partial Fractions, Parametric)

How to Solve: Integration (Basic, By Substitution, By Parts, Partial Fractions, Parametric)

GCSE / A-Level Maths

Introduction

"Mastering integration doesn’t just get you 10–15 marks on your A-Level exam—it’s the key to calculating areas under curves, modelling real-world motion, and even predicting future trends in physics and economics. One question could be the difference between a B and an A!

What You Need To Know First

1. Differentiation rules – You must recognise derivatives of common functions (e.g., d/dx of x² is 2x).
2. Algebraic manipulation – Rearranging equations, factorising, and simplifying fractions.
3. Basic integration – The reverse of differentiation (e.g., ∫xⁿ dx = xⁿ⁺¹/(n+1) + C).

Key Vocabulary

Formulas To Know

Step-by-Step Method

1. Basic Integration (Reverse of Differentiation)

Steps: 1. Identify the function type (polynomial, trig, exponential, etc.). 2. Apply the correct rule (e.g., power rule, trig rule). 3. Add +C (for indefinite integrals). 4. Simplify if needed.

Example: ∫(3x² + 4x – 5) dx 1. Split into terms: ∫3x² dx + ∫4x dx – ∫5 dx 2. Apply power rule:
- 3x³/3 = x³
- 4x²/2 = 2x²
- 5x 3. Add C: x³ + 2x² – 5x + C

2. Integration by Substitution

Steps: 1. Choose u (usually the inner function or the part whose derivative is present). 2. Find du/dx and rearrange to dx = du/(du/dx). 3. Rewrite the integral in terms of u. 4. Integrate with respect to u. 5. Substitute back to x. 6. Add +C.

Example: ∫2x(x² + 1)³ dx 1. Let u = x² + 1 (inner function). 2. du/dx = 2x → dx = du/(2x). 3. Rewrite: ∫2x u³ (du/(2x)) = ∫u³ du. 4. Integrate: u⁴/4 + C. 5. Substitute back: (x² + 1)⁴/4 + C.

3. Integration by Parts

Steps: 1. Identify u and dv (use LIATE rule: Logs, Inverse trig, Algebraic, Trig, Exponential). 2. Differentiate u to get du. 3. Integrate dv to get v. 4. Apply the formula: ∫u dv = uv – ∫v du. 5. Simplify and integrate the remaining term. 6. Add +C.

Example: ∫x eˣ dx 1. Let u = x (algebraic), dv = eˣ dx. 2. du = dx, v = eˣ. 3. Apply formula: x eˣ – ∫eˣ dx. 4. Integrate: x eˣ – eˣ + C. 5. Factor: eˣ(x – 1) + C.

4. Partial Fractions

Steps: 1. Factorise the denominator (if possible). 2. Write the partial fraction decomposition (e.g., A/(x+a) + B/(x+b)). 3. Multiply through by the denominator to eliminate fractions. 4. Solve for A and B by substituting values of x or equating coefficients. 5. Rewrite the integral using the partial fractions. 6. Integrate each term separately. 7. Add +C.

Example: ∫1/((x+1)(x+2)) dx 1. Decompose: 1/((x+1)(x+2)) = A/(x+1) + B/(x+2). 2. Multiply: 1 = A(x+2) + B(x+1). 3. Solve:
- Let x = -1: 1 = A(1) → A = 1.
- Let x = -2: 1 = B(-1) → B = -1. 4. Rewrite: ∫(1/(x+1) – 1/(x+2)) dx. 5. Integrate: ln|x+1| – ln|x+2| + C. 6. Simplify: ln|(x+1)/(x+2)| + C.

5. Parametric Integration

Steps: 1. Express y and dx in terms of t (the parameter). 2. Rewrite the integral as ∫y (dx/dt) dt. 3. Integrate with respect to t. 4. Substitute back to x and y if needed. 5. Add +C.

Example: Given x = t², y = t³, find ∫y dx from t=0 to t=1. 1. dx/dt = 2t. 2. Rewrite: ∫t³ (2t) dt = ∫2t⁴ dt. 3. Integrate: 2t⁵/5. 4. Evaluate from 0 to 1: 2/5 – 0 = 2/5.

Worked Examples

Example 1 – Basic

Question: ∫(5x⁴ – 3x² + 2) dx Solution: 1. Split: ∫5x⁴ dx – ∫3x² dx + ∫2 dx. 2. Apply power rule:
- 5x⁵/5 = x⁵
- 3x³/3 = x³
- 2x 3. Add C: x⁵ – x³ + 2x + C. What we did and why: We split the integral into simpler terms and applied the power rule to each. Always check for constants and signs!

Example 2 – Medium (Substitution)

Question: ∫x e^(x²) dx Solution: 1. Let u = x² (inner function). 2. du/dx = 2x → dx = du/(2x). 3. Rewrite: ∫x eᵘ (du/(2x)) = ½ ∫eᵘ du. 4. Integrate: ½ eᵘ + C. 5. Substitute back: ½ e^(x²) + C. What we did and why: We spotted that x is the derivative of x², making substitution the best method. Always look for du/dx in the integral!

Example 3 – Exam-Style (By Parts + Trig)

Question: ∫x sin(2x) dx Solution: 1. Let u = x (algebraic), dv = sin(2x) dx. 2. du = dx, v = -½ cos(2x). 3. Apply formula: -½ x cos(2x) – ∫-½ cos(2x) dx. 4. Simplify: -½ x cos(2x) + ½ ∫cos(2x) dx. 5. Integrate: -½ x cos(2x) + ¼ sin(2x) + C. What we did and why: We used integration by parts (LIATE rule) and remembered to integrate dv correctly. Always double-check signs!

Common Mistakes

Exam Traps

1-Minute Recap

"Right, listen up—this is your last-minute integration survival guide. For basic integrals, just reverse differentiation and add +C. Substitution? Pick u as the inner function and make sure du/dx is in the integral. By parts? Use LIATE to choose u and dv, then apply uv – ∫v du. Partial fractions? Factorise the denominator, split into simpler fractions, and solve for constants. Parametric? Rewrite dx as (dx/dt) dt and integrate with respect to t. And whatever you do, don’t forget +C! Now go smash that exam!