Antiderivative — definition, family of solutions (+C): Worked examples

The scenario matrix

Every antiderivative problem you will meet falls into one of these cells. The rightmost column names the example that clears it.

Cell	What makes it tricky	Example
A. Plain power, $n \geq 0$	just run the power rule forward	Ex 1
B. Negative power, $n \neq = - 1$	still power rule — but sign of exponent flips	Ex 2
C. The forbidden $n = - 1$	power rule divides by zero → logarithm	Ex 3
D. Disconnected domain	a gap means two constants, not one	Ex 3
E. Particular solution	an initial condition pins one curve	Ex 4
F. Sum / constant multiple	linearity: break it apart	Ex 5
G. Trig + degenerate check	sign traps; verify at special angles	Ex 6
H. Real-world word problem	velocity → position, units matter	Ex 7
I. Exam twist (rewrite first)	looks un-integrable until algebra	Ex 8

We will hit A B C D E F G H I — every cell.

Ex 1 — Cell A: plain power

This is the clean Power Rule (Integration) with $n = 4$ : $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ .

Ex 2 — Cell B: negative power (but not $- 1$ )

Ex 3 — Cells C & D: the forbidden power and the split domain

Worked example Statement

Find $\int \frac{1}{x} d x$ , and give the fully general family.

Forecast: try the power rule with $n = - 1$ . What goes wrong the instant you divide by $n + 1$ ?

Step 1. Attempt the power rule: $n + 1 = - 1 + 1 = 0$ , so the formula demands $\frac{x ^{0}}{0}$ — division by zero. Why this step? Seeing exactly where the rule detonates tells us we need a different tool. The question "what function has slope $\frac{1}{x}$ ?" is answered not by a power but by the natural logarithm.

Step 2. Recall $\frac{d}{d x} ln x = \frac{1}{x}$ for $x > 0$ . But $\frac{1}{x}$ also exists for $x < 0$ , where $ln x$ is undefined. The fix is the absolute value: $\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x} (x \neq = 0) .$ Why this step? $∣ x ∣$ lets one formula cover both signs of $x$ (see figure — the two branches).

Step 3 (Cell D). The domain $x \neq = 0$ is two disconnected intervals. The "no-other-antiderivatives" theorem (from the Mean Value Theorem) needs one connected interval, so each branch gets its own constant: $\int \frac{1}{x} d x = {ln (- x) + C_{1}, ln (x) + C_{2}, x < 0 x > 0.$ Why this step? Nothing forces the left branch to sit at the same height as the right — there is a gap between them, so they are independent.

Verify: for $x > 0$ , $\frac{d}{d x} ln x = \frac{1}{x}$ ✓. For $x < 0$ , $\frac{d}{d x} ln (- x) = \frac{- 1}{- x} = \frac{1}{x}$ ✓.

Ex 4 — Cell E: particular solution from an initial condition

Ex 5 — Cell F: sum and constant multiple (linearity)

Worked example Statement

Find $\int (3 x^{2} + 5 cos x - \frac{2}{x ^{2}}) d x$ .

Forecast: integration lets you attack each term separately and pull constants out front. Predict the three pieces before checking.

Step 1. Split by linearity: $\int 3 x^{2} d x + \int 5 cos x d x - \int 2 x^{- 2} d x .$ Why this step? Differentiation is linear (derivative of a sum is the sum of derivatives), so its reverse is too. This turns one hard-looking integral into three easy ones.

Step 2. Do each piece:

$\int 3 x^{2} d x = 3 \cdot \frac{x ^{3}}{3} = x^{3}$ .
$\int 5 cos x d x = 5 sin x$ (since $\frac{d}{d x} sin x = cos x$ ).
$\int 2 x^{- 2} d x = 2 \cdot \frac{x ^{- 1}}{- 1} = - 2 x^{- 1} = - \frac{2}{x}$ ; with the leading minus it becomes $+ \frac{2}{x}$ .

Step 3. Recombine with a single $+ C$ (one constant covers the whole sum, on a connected domain): $\int (3 x^{2} + 5 cos x - \frac{2}{x ^{2}}) d x = x^{3} + 5 sin x + \frac{2}{x} + C .$

Verify: $\frac{d}{d x} (x^{3} + 5 sin x + \frac{2}{x} + C) = 3 x^{2} + 5 cos x - \frac{2}{x ^{2}}$ ✓.

Ex 6 — Cell G: trig with a degenerate-angle sanity check

Ex 7 — Cell H: real-world word problem (velocity → position)

Worked example Statement

A ball is thrown upward. Its velocity (metres per second) is $v (t) = 20 - 10 t$ . Its height at $t = 0$ is $s (0) = 2$ m. Find the height $s (t)$ , and where the ball is at its peak.

Forecast: velocity is the slope of position, so position is an antiderivative of velocity. Guess: when is velocity zero (the peak)?

Step 1. Position is the antiderivative of velocity: $s (t) = \int (20 - 10 t) d t = 20 t - 10 \cdot \frac{t ^{2}}{2} + C = 20 t - 5 t^{2} + C .$ Why this step? $s^{'} (t) = v (t)$ by definition of velocity, so recovering $s$ from $v$ is exactly antidifferentiation. Units: $[20 t]$ is $\frac{m}{s} \cdot s = m$ ✓; $[5 t^{2}]$ is $\frac{m}{s ^{2}} \cdot s^{2} = m$ ✓.

Step 2. Use $s (0) = 2$ : $20 (0) - 5 (0)^{2} + C = 2 \Rightarrow C = 2$ . So $s (t) = 20 t - 5 t^{2} + 2.$ Why this step? The initial height is the one piece of info that fixes $C$ — the ball's launch height.

Step 3. Peak is where velocity $= 0$ : $20 - 10 t = 0 \Rightarrow t = 2$ s. Height there: $s (2) = 20 (2) - 5 (2)^{2} + 2 = 40 - 20 + 2 = 22 m .$ Why this step? At the top, the ball momentarily stops rising ( $v = 0$ ), so $s$ has zero slope — a maximum.

Verify: $s^{'} (t) = 20 - 10 t = v (t)$ ✓; $s (0) = 2$ ✓; peak height $22$ m at $t = 2$ ✓.

Ex 8 — Cell I: exam twist (rewrite before you can integrate)

Recall Quick self-test (cover the answers)

Which cell fails the power rule, and what replaces it? ::: Cell C, $n = - 1$ ; replaced by $ln ∣ x ∣ + C$ . In Ex 4, what is $C$ ? ::: $C = 12$ . In Ex 7, the peak height and time? ::: $22$ m at $t = 2$ s. Why two constants in Ex 3? ::: The domain $x \neq = 0$ is two disconnected intervals; each gets its own constant. First move on Ex 8's quotient? ::: Split the fraction into $x + x^{- 1/2}$ before integrating.

Connections

Power Rule (Integration) — Ex 1, 2, 8 and its $n = - 1$ failure in Ex 3.
Logarithmic & Exponential Integrals — the $ln ∣ x ∣$ of Ex 3.
Mean Value Theorem — why one interval means one constant (Ex 3, 5).
Differential Equations — Initial Value Problems — Ex 4, 6, 7 fixing $C$ .
Fundamental Theorem of Calculus / Definite Integral — where these families let $+ C$ cancel.