# Chapter 6 Revision: Indefinite Integration

F.5 Mathematics Module 1 · Chapter 6 Indefinite Integration · Lesson 3

The whole chapter on one card: the basic integrals, when to rewrite and when to substitute, and how a point or a starting value fixes the constant.

## Key points

- **Indefinite integral ／ 不定積分**: integration is the reverse of differentiation. If $\frac{d}{dx}F(x) = f(x)$, then $\int f(x)\,dx = F(x) + C$. Check any answer by differentiating it.
- **Exam language**: "Hence" means use the previous part; "Hence, or otherwise" also accepts any other correct method. A "hence" part (a) often gives $F(x)$.
- **Rewrite first**: expand, divide or split the fraction into simpler terms, then integrate term by term. Never integrate a product factor by factor.
- **Linear inside** $ax + b$: divide by the coefficient of $x$, e.g. $\int e^{2x}\,dx = \tfrac{1}{2}e^{2x} + C$.
- **Integration by substitution ／ 代換積分法**: let $u = g(x)$, change $dx$ to $du$ as well with $du = g'(x)\,dx$, integrate in $u$, then put $g(x)$ back.
- **Bars**: write $+\,C$ and the bars in $\ln\lvert\ldots\rvert$; drop the bars only when the inside is always positive, e.g. $\ln(x^2 + 8)$.
- **An $x$ left over**: let $u$ = the inside, then write $x$ in terms of $u$: if $u = x + 3$, then $x = u - 3$.
- **Curve from its slope**: $y = \int \frac{dy}{dx}\,dx$; a point on the curve gives the **arbitrary constant ／ 任意常數** $C$. A turning point gives $\frac{dy}{dx} = 0$.
- **Rate of change ／ 變化率**: if $\frac{dQ}{dt} = R(t)$, then $Q = \int R(t)\,dt$; the starting value $Q(0)$ gives $C$.

## Must know

| Idea | Rule |
|------|------------------|
| Reverse of differentiation | If $\dfrac{d}{dx}F(x) = f(x)$, then $\int f(x)\,dx = F(x) + C$ |
| Basic integrals | $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ $(n \ne -1)$; $\int \dfrac{1}{x}\,dx = \ln\lvert x\rvert + C$; $\int e^x\,dx = e^x + C$ |
| Linear inside | $\int (ax+b)^n\,dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + C$ $(n \ne -1)$; $\int e^{ax+b}\,dx = \dfrac{1}{a}e^{ax+b} + C$ |
| Substitution | $u = g(x)$, $du = g'(x)\,dx$: $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$, then put $g(x)$ back |
| Derivative on top | $\int \dfrac{g'(x)}{g(x)}\,dx = \ln\lvert g(x)\rvert + C$ |
| Second derivative given | From $\dfrac{d^2y}{dx^2}$ integrate twice; a turning point gives $\dfrac{dy}{dx} = 0$ for $C_1$ |
| Rate of change | If $\dfrac{dQ}{dt} = R(t)$, then $Q = \int R(t)\,dt$; $Q(0)$ gives $C$ |

Not sure of an answer? Differentiate it: you must get the integrand back.

## Common mistakes

- ✗ $\int x^{-1}\,dx = \dfrac{x^0}{0} + C$. ✓ The power rule needs $n \ne -1$: $\int x^{-1}\,dx = \ln\lvert x\rvert + C$.
- ✗ $\int x(x^2 + 4)\,dx = \int x\,dx \cdot \int (x^2 + 4)\,dx$. ✓ Never integrate a product factor by factor. Expand first: $\int (x^3 + 4x)\,dx = \tfrac{1}{4}x^4 + 2x^2 + C$.
- ✗ $\int e^{2x}\,dx = e^{2x} + C$. ✓ With a linear inside, divide by the coefficient of $x$: $\tfrac{1}{2}e^{2x} + C$.
- ✗ $\int \dfrac{1}{x + 5}\,dx = \ln(x + 5)$. ✓ Write $+\,C$ and the bars: $\ln\lvert x + 5\rvert + C$. Drop the bars only when the inside is always positive, e.g. $\ln(x^2 + 8)$.
- ✗ $\int (e^{2x} - 3)^2\,dx = \dfrac{(e^{2x} - 3)^3}{3} + C$. ✓ The factor $2e^{2x}$, the derivative of the inside, is missing. Expand instead: $\tfrac{1}{4}e^{4x} - 3e^{2x} + 9x + C$.

## Quick check

1. Find $\int \dfrac{x^2 + 3}{x}\,dx$.
2. Find $\int (4x - 1)^5\,dx$.
3. Find $\int \dfrac{x^2}{x^3 + 5}\,dx$.
4. Find $\int x\sqrt{x + 3}\,dx$.
5. At any point $(x, y)$ of a curve, $\frac{d^2y}{dx^2} = 6x + 2$, and the minimum point of the curve is $(1, 2)$. Find the equation of the curve.

## Answers

1. Divide first: $\int \left(x + \dfrac{3}{x}\right)dx = \dfrac{x^2}{2} + 3\ln\lvert x\rvert + C$.
2. Linear inside, $a = 4$: $\dfrac{(4x - 1)^6}{24} + C$.
3. $u = x^3 + 5$, $du = 3x^2\,dx$: $\tfrac{1}{3}\ln\lvert x^3 + 5\rvert + C$.
4. $u = x + 3$, $x = u - 3$: $\int (u^{3/2} - 3u^{1/2})\,du = \tfrac{2}{5}(x + 3)^{5/2} - 2(x + 3)^{3/2} + C$.
5. $\frac{dy}{dx} = 3x^2 + 2x + C_1$, $0 = 3 + 2 + C_1$, $C_1 = -5$; $y = x^3 + x^2 - 5x + C_2$, $2 = 1 + 1 - 5 + C_2$, $C_2 = 5$: $y = x^3 + x^2 - 5x + 5$.

## Did you know?

Computer algebra software can often tell whether an integral has an answer in the usual functions, using a method based on Robert Risch's 1969 work; for $\dfrac{1}{\ln x}$ there is no such answer.