Integration Techniques: Complete Guide
Unlike differentiation — which has systematic rules for all functions — integration requires recognizing which technique applies to each type of integrand. The main techniques are u-substitution (the chain rule in reverse), integration by parts (the product rule in reverse), trigonometric substitution, and partial fraction decomposition.
This guide explains each technique, the integrands they handle, and worked examples. Developing pattern recognition for "which technique?" is the core skill of integral calculus.
Formula
∫u dv = uv − ∫v du | ∫f(g)g'dx = ∫f(u)du (u-sub)
U-Substitution (Reverse Chain Rule)
When to use: the integrand is a composite function f(g(x))·g'(x) — the derivative of the inner function appears as a factor.
Steps: (1) Identify u = g(x) — the inner function. (2) Compute du = g'(x)dx. (3) Rewrite the integral entirely in terms of u. (4) Integrate. (5) Substitute back.
Example: ∫2x·cos(x²)dx. Let u = x², du = 2x dx. Integral becomes ∫cos(u)du = sin(u) + C = sin(x²) + C.
Example: ∫x²(x³+1)⁵dx. Let u = x³+1, du = 3x²dx → x²dx = du/3. Integral: (1/3)∫u⁵du = u⁶/18 + C = (x³+1)⁶/18 + C.
| Integrand Pattern | Substitution | Result Form |
|---|---|---|
| ∫f(ax+b)dx | u = ax+b | (1/a)F(ax+b) + C |
| ∫xⁿ·f(xⁿ⁺¹)dx | u = xⁿ⁺¹ | (1/(n+1))F(u) + C |
| ∫f'(x)/f(x)dx | u = f(x) | ln|f(x)| + C |
| ∫sin(x)·cos(x)dx | u = sin x or cos x | (1/2)sin²x + C or −(1/2)cos²x + C |
| ∫eᶠ⁽ˣ⁾·f'(x)dx | u = f(x) | eᶠ⁽ˣ⁾ + C |
Integration by Parts
Formula: ∫u dv = uv − ∫v du. Derived from the product rule in reverse.
When to use: products of different function types — polynomial × exponential, polynomial × trig, polynomial × log, inverse trig alone.
LIATE rule for choosing u: Logarithmic > Inverse trig > Algebraic (polynomial) > Trigonometric > Exponential. Choose u as the type that appears first in LIATE; dv is everything else.
Example: ∫x·eˣdx. u = x, dv = eˣdx → du = dx, v = eˣ. Result: x·eˣ − ∫eˣdx = x·eˣ − eˣ + C = eˣ(x−1) + C.
Example: ∫ln(x)dx. u = ln x, dv = dx → du = (1/x)dx, v = x. Result: x·ln x − ∫x·(1/x)dx = x·ln x − x + C.
Repeated application: ∫x²·eˣdx requires two rounds of integration by parts. After the first: x²eˣ − 2∫x·eˣdx. The second integral was solved above. Total: x²eˣ − 2xeˣ + 2eˣ + C = eˣ(x²−2x+2) + C.
Partial Fraction Decomposition
When to use: rational functions (polynomial ÷ polynomial) where the degree of numerator < degree of denominator.
Process: factor the denominator, decompose into simpler fractions with undetermined coefficients, solve for coefficients by equating numerators, integrate each piece.
Example: ∫(5x+4)/[(x+1)(x+2)]dx. Decompose: (5x+4)/[(x+1)(x+2)] = A/(x+1) + B/(x+2). Multiply through: 5x+4 = A(x+2) + B(x+1). At x=−1: −1 = A(1) → A = −1. At x=−2: −6 = B(−1) → B = 6. Integral: −ln|x+1| + 6ln|x+2| + C.
For repeated factors like (x+1)²: decompose as A/(x+1) + B/(x+1)². For irreducible quadratics like (x²+1): decompose as (Ax+B)/(x²+1).
Trigonometric Substitution
When to use: integrands containing √(a²−x²), √(a²+x²), or √(x²−a²).
Pattern matching: √(a²−x²) → substitute x = a·sin θ. √(a²+x²) → substitute x = a·tan θ. √(x²−a²) → substitute x = a·sec θ.
Example: ∫√(4−x²)dx. x = 2sin θ, dx = 2cos θ dθ, √(4−x²) = 2cos θ. Integral: ∫2cos θ · 2cos θ dθ = 4∫cos²θ dθ = 2∫(1+cos 2θ)dθ = 2θ + sin 2θ + C. Back-substitute: θ = arcsin(x/2), sin 2θ = 2sin θ cos θ = 2(x/2)(√(4−x²)/2) = x√(4−x²)/2. Result: 2arcsin(x/2) + x√(4−x²)/2 + C.