# Chapter 18: Integration Techniques, Part One

> The workhorses of integration: u-substitution (indefinite and with changed limits), the inside-function rule, integration by parts with LIATE, and partial fractions — distinct, repeated and irreducible-quadratic factors — capped by one big worked example.

**Canonical:** https://calculus.kitzo.work/chapter/integration-1  
**Level:** Core  
**Estimated time:** 55 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- u-substitution
- Substitution with limits
- Choosing u
- Integration by parts
- LIATE
- Partial fractions
- ∫ln x dx

**Why this matters**

Differentiation 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: **L**ogarithmic, **I**nverse trig, **A**lgebraic, **T**rig, **E**xponential. 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 out**

Partial 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.

## Practice exercises

1. What is the right choice of u for ∫ x²(x³ + 1)ⁿ dx?

- A. u = x²
- B. u = x³ + 1
- C. u = (x³ + 1)ⁿ
- D. u = x

2. Evaluate ∫₀² 2x(x² + 1) dx using substitution.

3. Write ∫ dx/(x − 3) dx. (Include + C.)

4. For ∫ x·ln x dx, LIATE says choose:

- A. u = x, dv = ln x dx
- B. u = ln x, dv = x dx
- C. u = x ln x, dv = dx
- D. no parts needed

5. Evaluate ∫₁ᵉ (1/x) dx.

6. Evaluate ∫ x·eˣ dx. (Include + C.)

7. Decompose 2x/((x−1)(x+1)) = A/(x−1) + B/(x+1). Then A equals:

- A. 2
- B. 0
- C. 1
- D. −1

8. Evaluate ∫₀¹ 3x²(x³ + 1)⁴ dx using substitution.

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