Basic integration rules — power, trig, exponential, log: Worked examples

We only need anti-differentiation ("what did I differentiate to get this?") and linearity (split sums, pull out constants). Every symbol below was earned in the parent; where a new idea appears I re-build it here.

The scenario matrix

Think of every basic integral as landing in one cell of this table. If we cover one worked example per cell, we cover everything this topic can throw.

Cell	Case class	What makes it tricky	Hit by Example
A	Positive integer power	plain power rule	1
B	Fractional / root power	rewrite root as exponent first	1
C	Negative power ( $n \leq - 2$ )	the sign of $n + 1$ flips	2
D	The forbidden $n = - 1$	power rule dies, log takes over	3
E	Trig with the minus-sign trap	$\int sin = - cos$	4
F	Exponential, base $e$ vs base $a$	divide by $ln a$ only for base $a$	5
G	Zero / degenerate input	integrand $= 0$ , or a constant	6
H	Real-world word problem	speed → position, choose $+ C$	7
I	Exam twist: hidden power	algebra before you can integrate	8
J	Limiting/edge behaviour	what happens as $n \to - 1$	9

Below, each example names the cell(s) it clears.

The figure below draws this same matrix as a coloured grid — each pastel block is one cell, and the colour just groups related cases (purple = power-type, coral = the tricky negative/log region, mint = trig & exponential, butter = degenerate/word-problem). Use it as a checklist: every block gets cleared by exactly one worked example.

Figure — Basic integration rules — power, trig, exponential, log

Worked examples

Worked example Example 1 — Cells A & B: integer and root powers together

$\int (x^{3} + 3 x) d x .$

Forecast: raise each power by one and divide. Guess the cube root becomes something like $x^{4/3}$ — but with what fraction in front?

Rewrite the root as a power. $3 x = x^{1/3}$ . Why this step? The power rule only recognises the form $x^{n}$ ; a root symbol is invisible to it. Turning $3 x$ into $x^{1/3}$ makes $n = \frac{1}{3}$ visible. Domain note. The real cube root $3 x$ is defined for all $x$ (including negatives), but the fractional-power symbol $x^{1/3}$ is only unambiguous for $x \geq 0$ . So we do the rewrite — and the final answer — on the domain $x \geq 0$ ; for $x < 0$ interpret $x^{1/3}$ as the real cube root and the formula still differentiates back correctly.
Apply the power rule to $x^{3}$ . Here $n = 3$ , so $n + 1 = 4$ : $\frac{x ^{4}}{4}$ . Why this step? Differentiation drops the power by one, so to run it backwards we raise the power and divide by the new power to cancel the leftover factor.
Apply the power rule to $x^{1/3}$ . Here $n + 1 = \frac{1}{3} + 1 = \frac{4}{3}$ : $\frac{x ^{4/3}}{4/3} = \frac{3}{4} x^{4/3}$ . Why this step? Dividing by a fraction $\frac{4}{3}$ is multiplying by its reciprocal $\frac{3}{4}$ — that reciprocal is where the $\frac{3}{4}$ out front comes from.
Add $+ C$ . $\frac{x ^{4}}{4} + \frac{3}{4} x^{4/3} + C .$

Verify: differentiate. $\frac{d}{d x} \frac{x ^{4}}{4} = x^{3}$ ✓, and $\frac{d}{d x} \frac{3}{4} x^{4/3} = \frac{3}{4} \cdot \frac{4}{3} x^{1/3} = x^{1/3} = 3 x$ ✓.

Worked example Example 3 — Cell D: the forbidden

$n = - 1$ $\int \frac{- 2}{x} d x .$

Forecast: this looks like a power with $n = - 1$ . But the power rule would divide by $n + 1 = 0$ . Does that mean the integral doesn't exist? (No — a different function saves us.)

Recognise $n = - 1$ . $\frac{- 2}{x} = - 2 x^{- 1}$ , so $n = - 1$ and $n + 1 = 0$ . Why this step? Spotting $n = - 1$ is the whole game — it is the one value the power rule forbids because you cannot divide by zero.
Switch to the log rule. $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ (this is the function that plugs the hole). Why this step? We need a function whose derivative is $\frac{1}{x}$ . From Natural log and exponential functions, $\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x}$ for all $x \neq = 0$ — even negative $x$ , which is why the absolute value $∣ x ∣$ is there.
Pull the constant out. $- 2 ln ∣ x ∣ + C .$

Verify: $\frac{d}{d x} (- 2 ln ∣ x ∣) = - 2 \cdot \frac{1}{x} = \frac{- 2}{x}$ ✓, valid for every $x \neq = 0$ .

Worked example Example 4 — Cell E: the sin/cos minus-sign trap

$\int (4 cos x - 3 sin x) d x .$

Forecast: one of these picks up a minus sign when integrated, and it is not the one you might expect. Which term flips?

Integrate $4 cos x$ . Since $\frac{d}{d x} sin x = cos x$ , we get $4 sin x$ . Why this step? Read the derivative backwards — cosine is the derivative of sine, so sine is the antiderivative of cosine. No sign change.
Integrate $- 3 sin x$ . We need $F^{'} = sin x$ . We know $\frac{d}{d x} cos x = - sin x$ , the negative of what we want, so $\int sin x d x = - cos x$ . Thus $- 3 \cdot (- cos x) = + 3 cos x$ . Why this step? The minus lives on the cosine's derivative, so it re-appears here — and combined with the $- 3$ out front, the two minuses cancel into a plus.
Combine. $4 sin x + 3 cos x + C .$

Verify: $\frac{d}{d x} (4 sin x + 3 cos x) = 4 cos x - 3 sin x$ ✓ — exactly the integrand. See Trigonometric integrals for the extended table.

The figure below plots $sin x$ , $cos x$ , and the dashed $- cos x$ so you can see where the minus sign is born: integrating $cos x$ lands on $sin x$ with no flip, but integrating $sin x$ lands on the dashed curve $- cos x$ , sitting below the axis exactly where you might expect $+ cos x$ .

Worked example Example 7 — Cell H: real-world word problem (velocity → position)

Set up the coordinate system first. Let $s (t)$ be the ball's height above the ground in metres, measured upward as positive, with $t$ in seconds. The ball is dropped from rest at height $s (0) = 20$ m and accelerates downward under gravity, so its velocity (upward-positive) is $v (t) = - 9.8 t$ — negative because the ball moves downward. Find $s (t)$ .

Forecast: position is the antiderivative of velocity. Integrating $- 9.8 t$ gives a $t^{2}$ term with a leading number — and a $+ C$ that we must pin down using the starting height.

Integrate velocity to get position. $s (t) = \int v (t) d t = \int - 9.8 t d t = - 9.8 \cdot \frac{t ^{2}}{2} + C = - 4.9 t^{2} + C .$ Why this step? Velocity is the derivative of position ( $v = d s / d t$ ), so position is the antiderivative of velocity. Power rule with $n = 1$ gives $\frac{t ^{2}}{2}$ , and $- 9.8/2 = - 4.9$ . The same sign convention (upward positive) is used throughout, so the $- 4.9$ stays negative.
Pin down $C$ with the initial condition. The $+ C$ is the starting height (that's the "car's unknown start" from the parent's detective analogy). Substitute $t = 0$ and $s (0) = 20$ : $s (0) = - 4.9 (0)^{2} + C = C ⟹ C = 20.$ Why this step? We have one unknown ( $C$ ) and one known fact ( $s (0) = 20$ ); solving the equation $C = 20$ fixes it. (Here $C$ came out equal to the height itself, $20$ , because the $t^{2}$ term vanishes at $t = 0$ .)
Write the position function. $s (t) = 20 - 4.9 t^{2} .$

Verify: $s (0) = 20 - 4.9 (0)^{2} = 20$ ✓ (units: metres). Differentiate: $\frac{d s}{d t} = \frac{d}{d t} (20 - 4.9 t^{2}) = - 9.8 t = v (t)$ ✓ (units: m/s, negative = downward, consistent). At $t = 2$ : $s = 20 - 4.9 (4) = 0.4$ m — nearly landed. See Definite integrals & FTC for the version with limits of integration.

Worked example Example 8 — Cell I: exam twist, hidden power (algebra first)

$\int \frac{x ^{2} + 3 x - 1}{x} d x .$

Forecast: there is no quotient rule for integration. You must divide through first so each term becomes a plain power. Predict the three exponents you'll get.

Split the fraction over $x = x^{1/2}$ . $\frac{x ^{2}}{x ^{1/2}} + \frac{3 x}{x ^{1/2}} - \frac{1}{x ^{1/2}} = x^{3/2} + 3 x^{1/2} - x^{- 1/2}$ . Why this step? Subtract exponents when dividing like bases ( $2 - \frac{1}{2} = \frac{3}{2}$ , $1 - \frac{1}{2} = \frac{1}{2}$ , $0 - \frac{1}{2} = - \frac{1}{2}$ ). This converts a "product/quotient" into a clean sum of powers — the only form the basic rules handle. (No product/quotient rule exists; that would need Integration by parts or Integration by substitution.)
Integrate each power.
- $\int x^{3/2} d x = \frac{x ^{5/2}}{5/2} = \frac{2}{5} x^{5/2}$
- $\int 3 x^{1/2} d x = 3 \cdot \frac{x ^{3/2}}{3/2} = 2 x^{3/2}$
- $\int - x^{- 1/2} d x = - \frac{x ^{1/2}}{1/2} = - 2 x^{1/2}$ Why this step? Straight power rule term by term; the messy fractions come from dividing by $n + 1$ .
Combine. $\frac{2}{5} x^{5/2} + 2 x^{3/2} - 2 x^{1/2} + C .$

Verify: $\frac{d}{d x} (\frac{2}{5} x^{5/2} + 2 x^{3/2} - 2 x^{1/2}) = x^{3/2} + 3 x^{1/2} - x^{- 1/2}$ , which recombines to $\frac{x ^{2} + 3 x - 1}{x}$ ✓.

Worked example Example 9 — Cell J: the limiting edge, why

$n = - 1$ is truly special Consider $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ and let $n$ creep toward $- 1$ . What breaks, and what takes over?

Forecast: as $n \to - 1$ the denominator $n + 1 \to 0$ , so the formula blows up — yet the actual integral $\int x^{- 1} d x$ is perfectly finite ( $ln ∣ x ∣$ ). How do both facts live together?

Look at the denominator. As $n \to - 1$ , $n + 1 \to 0$ , so $\frac{x ^{n + 1}}{n + 1}$ has a $\frac{1}{0}$ — the power-rule expression genuinely diverges for a fixed nonzero $x \neq = 1$ . Why this step? It shows the exclusion $n \neq = - 1$ is not a technicality bolted on — the formula literally cannot be evaluated there.
See what fills the gap. The correct antiderivative at $n = - 1$ is a different species of function, $ln ∣ x ∣$ , not any power of $x$ . Why this step? Powers can never produce $\frac{1}{x}$ upon differentiation (their derivatives are again powers), so the logarithm — an entirely new function — is required.
Conclusion. $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C (n \neq = - 1)$ , and exactly at the excluded point, $\int x^{- 1} d x = ln ∣ x ∣ + C$ .

Verify (numeric peek at the blow-up): with $x = 2$ , evaluate $\frac{x ^{n + 1}}{n + 1}$ at $n = - 0.9$ and $n = - 0.99$ : you get about $1.03$ then $0.100$ times ten — more precisely $\approx 10.3$ and $\approx 103$ — growing without bound as $n \to - 1$ , confirming the divergence.

The figure below plots that expression against $n$ : watch the purple curve rocket upward as it nears the coral dashed line at $n = - 1$ , while the mint dashed line marks the finite value $ln 2$ that the true integral takes at the excluded point.

Recall One-line reason for each cell

Positive power ::: raise & divide, no sign fuss. Negative power ::: raise & divide, but $n + 1$ may be negative → a minus sign appears. $n = - 1$ ::: power rule dies (÷0); use $ln ∣ x ∣$ . $\int sin x$ ::: $- cos x$ (minus lives on cosine's derivative). $\int a^{x}$ ::: divide by $ln a$ ; base $e$ needs no divisor. Constant $k$ ::: $\int k d x = k x + C$ (it's $k x^{0}$ ). Word problem ::: $+ C$ is the initial condition — solve for it. Quotient/product integrand ::: rewrite into a sum of powers first; no quotient rule.

Connections

Parent: Basic integration rules
Derivatives — basic rules (every check here differentiates the answer)
Integration by substitution (when the integrand is a composite, not a plain power)
Integration by parts (when a genuine product resists the tricks above)
Definite integrals & FTC (Example 7 with actual limits of integration)
Natural log and exponential functions (source of the $ln ∣ x ∣$ and $\frac{a ^{x}}{l n a}$ rules)
Trigonometric integrals (extends Example 4's table)