Example 1: Anti derivatives of $\cos 2x$⁠, $3x^2 + 4x^3$ and $\frac{1}{x}$ by inspection

Basics you need

Part (i): an anti derivative of $\cos 2x$

1. Find a derivative that produces $\cos 2x$

   The function whose derivative is a cosine is a sine. Try $\sin 2x$⁠. By the chain rule (inner $u = 2x$⁠, outer $\sin u$⁠):

   $$\frac{d}{dx}\sin 2x = \cos 2x \cdot \frac{d}{dx}(2x) = 2\cos 2x.$$
   This is the method of inspection: we search our memory of derivatives for one whose result has the right shape. $\sin 2x$ gives the right function, $\cos 2x$⁠, but with an unwanted factor 2 produced by the inner function $2x$⁠.
   Standard formula (ii) of Section 7.2: $\frac{d}{dx}\sin x = \cos x$⁠; chain rule from Chapter 5. Basics: the chain rule backwards

2. Remove the extra factor 2

   Divide both sides by 2. A constant factor can be moved inside the derivative:

   $$\cos 2x = \frac12\,\frac{d}{dx}(\sin 2x) = \frac{d}{dx}\left(\frac12 \sin 2x\right).$$

   So an anti derivative of $\cos 2x$ is $\frac12\sin 2x$.

   $\frac{d}{dx}(kF) = k\frac{d}{dx}F$ for any constant $k$⁠. Halving the guess halves its derivative, which turns $2\cos 2x$ into exactly $\cos 2x$⁠.
   Constant multiple rule for derivatives. Basics: integration undoes differentiation

3. Check: differentiate the answer

   $$\frac{d}{dx}\left(\frac12\sin 2x\right) = \frac12\cdot 2\cos 2x = \cos 2x. \ \checkmark$$

   This holds for every real $x$⁠. Adding any constant also gives an anti derivative, so the full family is $\int\cos 2x\,dx = \frac12\sin 2x + C$⁠.

   An anti derivative is defined by its derivative, so differentiating the answer is a complete check, not just a spot test.
   Lesson: integration as anti-differentiation

Part (ii): an anti derivative of $3x^2 + 4x^3$

1. Recognise each term as a derivative

   By the power rule $\frac{d}{dx}x^n = nx^{n-1}$⁠:

   $$\frac{d}{dx}x^3 = 3x^2, \qquad \frac{d}{dx}x^4 = 4x^3.$$
   Each term already carries the "right" coefficient: $3x^2$ is exactly what $x^3$ differentiates to, and $4x^3$ is exactly what $x^4$ differentiates to. No adjustment of constants is needed.
   Standard formula (i) of Section 7.2. Basics: powers of x and the power rule

2. Add them

   The derivative of a sum is the sum of the derivatives, so

   $$\frac{d}{dx}\left(x^3 + x^4\right) = 3x^2 + 4x^3.$$

   So an anti derivative of $3x^2 + 4x^3$ is $x^3 + x^4$, for every real $x$⁠.

   Because differentiation works term by term, we may find an anti derivative of each term separately and add them. For integrals this is Property (III) of Section 7.2.1, proved just before this example.
   Sum rule. Lesson: integration as anti-differentiation

3. Check: differentiate the answer

   $$\frac{d}{dx}(x^3 + x^4) = 3x^2 + 4x^3. \ \checkmark$$

   The full family is $x^3 + x^4 + C$⁠.

   Part (ii), Step 2 was already this check written the other way round; repeating it as the last step makes the habit explicit.

Part (iii): an anti derivative of $\frac1x$⁠, $x \ne 0$

1. The positive side, $x > 0$

   We know

   $$\frac{d}{dx}(\log x) = \frac1x, \qquad x > 0.$$

   So on $x > 0$⁠, $\log x$ is an anti derivative of $\frac1x$⁠.

   The power rule cannot help here: $\frac1x = x^{-1}$⁠, and $\frac{x^{n+1}}{n+1}$ with $n = -1$ would divide by zero. That is why formula (i) excludes $n = -1$⁠, and why the logarithm is needed.
   Derivative of $\log x$ (Chapter 5, Theorem 5). Basics: logarithms and 1/x

2. The negative side, $x < 0$

   For $x < 0$⁠, $\log x$ does not exist, but $-x > 0$⁠, so $\log(-x)$ does. By the chain rule (inner $u = -x$⁠, outer $\log u$⁠):

   $$\begin{aligned} \frac{d}{dx}\log(-x) &= \frac{1}{-x}\cdot\frac{d}{dx}(-x) \\ &= \frac{1}{-x}\cdot(-1) = \frac1x, \quad x < 0. \end{aligned}$$
   The question allows every $x \ne 0$⁠, including negative $x$⁠, so an answer that only works for $x > 0$ would be incomplete. The two minus signs cancel, so $\log(-x)$ has exactly the derivative we want on the negative side.
   Chain rule. Basics: the chain rule backwards

3. Combine the two sides with $|x|$

   Since $|x| = x$ for $x > 0$ and $|x| = -x$ for $x < 0$⁠, Part (iii), Steps 1 and 2 say together

   $$\frac{d}{dx}\log|x| = \frac1x, \qquad x \ne 0.$$

   So $\int \frac1x\,dx = \log|x|$ is an anti derivative of $\frac1x$, valid on $x > 0$ and on $x < 0$⁠.

   One formula, $\log|x|$⁠, covers both sides: it equals $\log x$ on the right of 0 and $\log(-x)$ on the left. This is standard formula (xii) of Section 7.2, which the book now proves.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

4. Check: differentiate the answer at a negative and a positive point

   At $x = -2$⁠: $\log|x| = \log(-x)$ near $-2$⁠, and its derivative is $\frac{1}{-2}$⁠, the value of $\frac1x$ there. At $x = 3$⁠: the derivative of $\log x$ is $\frac13$⁠. Both match $\frac1x$⁠. $\checkmark$

   The family of anti derivatives is written $\log|x| + C$⁠. (Strictly, because $x = 0$ splits the domain into two separate intervals, the constant may even be different on the two sides. The book does not discuss this, but its Note after the formula list reminds us that the interval on which a function is defined must be kept in mind.)

   The general statement in Part (iii), Step 3 already is the check; testing one point on each side shows concretely that both branches are needed and both work.
   Lesson: integration as anti-differentiation

Every fact used, and where it comes from