Chapter 18: Integration Techniques, Part One

Why this mattersDifferentiation was a small kit of rules. Integration is the art of running those rules backwards — and it takes weapons. This chapter arms you with the three heavy cannons: substitution, parts, and partial fractions.

1 · The chain rule in reverse

The chain rule says d/dx F(g(x)) = f(g(x))·g′(x). Read it backwards: an integrand that is a composite times the derivative of the inside is an antiderivative in disguise. That is all u-substitution is — the chain rule run in reverse.

2 · The u-sub recipe (indefinite)

Pick u = g(x) — the inside of the composite.
Compute du = g′(x)dx.
Rewrite the integral entirely in u; no x may survive.
Integrate, then substitute back u = g(x), and add + C.

Worked example Example 1 — ∫ x·√(x² + 1) dx.
Step 1: inside function u = x² + 1 (it is under the roof).
Step 2: du = 2x dx, so x dx = du/2.
Step 3: ∫ √(u)·(du/2) = ½ · (2/3)u^(3/2) = (1/3)u^(3/2).
Step 4: back-substitute: (1/3)(x² + 1)^(3/2) + C. Check by differentiating — ⅓·(3/2)(x²+1)^(1/2)·2x = x√(x²+1). ✓

Missing a variable factor? Only a constant may be rescued (the ½ above came from 2x dx = du). If no x dx factor exists, that substitution is hopeless — find another weapon.

3 · Substitution in definite integrals

Two routes, one destination. The slow route back-substitutes, then uses the old x-limits. The fast route — the one exams reward — changes the limits to u-values: x = a ⇒ u = g(a), x = b ⇒ u = g(b). No x survives.

Worked example Example 2 — ∫₀¹ 2x·e^(x²) dx with changed limits.
Step 1: u = x², du = 2x dx.
Step 2: limits: x = 0 ⇒ u = 0; x = 1 ⇒ u = 1.
Step 3: ∫₀¹ e^u du = e¹ − e⁰ = e − 1.
No back-substitution, no x left anywhere. Cleaner and faster.

4 · How to choose u

The golden rule: u is the inside function whose derivative is also present (up to a constant). Hunting list, in order of likelihood:

Inside of a composite: u = g(x) when g′(x) lurks in the integrand.
The exponent: u = x² under e^(x²); the trig argument: u = 3x for sin 3x.
The denominator: u = x² + 4 when dx/(x²+4) style appears.
The whole awkward chunk wrapped in a power or log.

If two candidates survive, pick the one whose derivative clears out everything else. A constant leftover (the ½ in Example 1) is fine — a leftover x is fatal.

5 · Integration by parts

The product rule, reversed, gives:

∫ u dv = uv − ∫ v du

You choose u and dv; du and v follow. What to differentiate? LIATE — differentiate the highest-priority type present: Logarithmic, Inverse trig, Algebraic, Trig, Exponential. Differentiate the L, integrate the E.

Worked example Example 3 — ∫ x·cos x dx, then ∫ ln x dx.
Part 1: LIATE — u = x (algebraic beats trig for u), dv = cos x dx. So du = dx, v = sin x.
∫ x cos x dx = x·sin x − ∫ sin x dx = x sin x + cos x + C.
Part 2: ∫ ln x dx has no dv in sight — manufacture one: u = ln x, dv = dx. Then du = dx/x, v = x.
∫ ln x dx = x·ln x − ∫ x·(dx/x) = x ln x − x + C. Logarithms are always differentiated, never integrated.

6 · Partial fractions

Any rational integrand (polynomial over polynomial, degree top < degree bottom) splits into digestible pieces:

Distinct linear factors: P(x)/((x+a)(x+b)) = A/(x+a) + B/(x+b).
Repeated factor (x+a)²: you need both A/(x+a) and B/(x+a)².
Irreducible quadratic x² + bx + c: the piece is (Bx + C)/(x² + bx + c), integrating to a log plus an arctan.

Solve for the constants by clearing denominators and matching coefficients — or evaluating at the roots. Integrating A/(x+a) gives A·ln|x+a|; the irreducible-quadratic piece splits into a log (from the x-term) and an arctan (from the constant term, via completing the square).

Worked example Example 4 (a big one) — ∫ (3x + 1)/((x+1)(x² + 1)) dx.
Step 1: decompose: (3x+1)/((x+1)(x²+1)) = A/(x+1) + (Bx+C)/(x²+1).
Step 2: 3x + 1 = A(x² + 1) + (Bx + C)(x + 1). Set x = −1: −2 = 2A ⇒ A = −1.
Step 3: expand: (A + B)x² + (B + C)x + (A + C) = 3x + 1 ⇒ A + B = 0, B + C = 3, A + C = 1. So B = 1, C = 2. ✓
Step 4: integrate pieces: −∫ dx/(x+1) + ∫ x·dx/(x²+1) + 2∫ dx/(x²+1)
= −ln|x+1| + ½·ln(x²+1) + 2·arctan x + C.
Two leftover shapes survived: a plain log piece and an arctan piece — that is what irreducible quadratics always give you.

Watch outPartial fractions only work when degree(top) < degree(bottom). If not, do polynomial long division first. And in definite integrals, change the limits to u-values as soon as you substitute — carrying original x-limits into a u-integral is the classic silent killer.

Chapter formula sheet u = g(x), du = g′(x)dx · ∫ f(g(x))g′(x)dx = ∫ f(u)du
∫ₐᵇ f(g(x))g′(x)dx = ∫_{g(a)}^{g(b)} f(u)du (change the limits!)
∫ u dv = uv − ∫ v du · LIATE: Logarithmic → Inverse trig → Algebraic → Trig → Exponential
∫ x eˣ dx = (x−1)eˣ + C · ∫ ln x dx = x ln x − x + C
A/(x+a) + B/(x+b) · repeated (x+a)² needs A/(x+a) + B/(x+a)²
(Bx + C)/(x²+bx+c): log piece + arctan piece
∫ dx/(x² + a²) = (1/a)·arctan(x/a) + C · ∫ dx/x = ln|x| + C

Common mistakes

Substituting u back into the limits — the limits must become u-values, or back-substitute fully first.
Forgetting the du factor (no x dx in sight means this u will not work).
In parts, choosing u = cos x over u = x — apply LIATE, not vibes.
Dropping the +C on indefinite integrals, or adding it to a definite one.
Partial fractions on a top-heavy fraction without long division first.
Missing the Bx + C numerator for a quadratic factor — it is never just A.