Basic integration rules — power, trig, exponential, log: Visual walkthrough

We will build these words from scratch, in this order: area under a curve → slope of a curve → "undoing" a slope → raising a power → the lost constant $+ C$ .

Step 1 — What is a slope, in one picture?

WHAT. Look at the curve $y = x^{2}$ (a smooth valley). Pick a point on it and lay a ruler so it just kisses the curve there — that ruler is the tangent line, and its tilt is the slope.

WHY this first. Integration is the reverse of taking slopes. We cannot reverse a machine we have never watched run forwards. So we look at slopes first.

PICTURE. In the figure, the burnt-orange curve is $y = x^{2}$ . At three points we drew the teal tangent rulers. Notice: on the left the ruler tilts down (negative slope), in the middle it is flat (slope $0$ ), on the right it tilts up (positive slope). The slope changes as you walk along.

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

The rule that tells you the slope of $x^{2}$ at every point is its derivative, written $\frac{d}{d x}$ : $\frac{d}{d x} x^{2} = old power 2 power dropped by 1 x .$

Here $\frac{d}{d x}$ means "the slope-finding machine acting on what follows," the $2$ is the old exponent brought down front, and $x^{1}$ is the exponent $2$ lowered by one. See Derivatives — basic rules for why this happens.

Step 2 — The pattern hidden in every derivative

WHAT. Test it on several powers and stack them: $\frac{d}{d x} x^{5} = 5 x^{4}, \frac{d}{d x} x^{3} = 3 x^{2}, \frac{d}{d x} x^{1} = 1 \cdot x^{0} = 1.$

$x^{5} \to$ the $5$ jumps to the front, the exponent becomes $5 - 1 = 4$ .
$x^{3} \to$ the $3$ jumps to the front, exponent becomes $2$ .
$x^{1} \to$ the $1$ jumps front, exponent becomes $0$ , and $x^{0} = 1$ (any nonzero number to the power $0$ is $1$ ).

WHY. If we can see the machine's two moves clearly going forwards, we can plan how to undo them going backwards. Undoing "drop the power by 1" must mean "raise the power by 1." Undoing "multiply by the old power" must mean "divide by something."

PICTURE. The staircase figure: each orange tile is a power $x^{n}$ ; the teal arrow leaving it (labelled "differentiate") lands one step lower and writes the multiplier on the arrow. Read the arrows backwards (the plum arrows) and you are integrating.

Step 3 — Guessing the antiderivative

WHAT. We want a function whose slope is exactly $x^{n}$ . From Step 2, differentiating lowers the power, so to end up at $x^{n}$ we must start one power higher: guess $F = x^{n + 1}$ .

WHY this guess. It is the only guess that survives Step 2's "power drops by 1" move: start at $n + 1$ , drop by one, land on $n$ .

PICTURE. Test the guess by differentiating it: $\frac{d}{d x} x^{n + 1} = old power to front (n + 1) x^{(n + 1) - 1} = (n + 1) x^{n} .$

$n + 1$ is the exponent we chose; it slides to the front as a multiplier.
$(n + 1) - 1 = n$ is the exponent after the drop — exactly the power we wanted.

So our guess gives $x^{n}$ times an unwanted factor $(n + 1)$ . The figure shows this as a target ( $x^{n}$ , plum) and our arrow overshooting it by the orange stretch-factor $(n + 1)$ .

Step 4 — Fixing the overshoot by dividing

WHAT. Divide the guess by that constant: $\frac{d}{d x} [\frac{x ^{n + 1}}{n + 1}] = \frac{( n + 1 ) x ^{n}}{n + 1} = x^{n} . ✓$

The $\frac{1}{n + 1}$ out front is a constant — differentiation carries constants along untouched.
The $(n + 1)$ produced by differentiating meets the $\frac{1}{n + 1}$ we placed there and they cancel to $1$ .
What remains is a clean $x^{n}$ — precisely the slope we ordered.

WHY. This is why the power rule says "raise the power, then divide by the new power." Each verb undoes one of differentiation's two moves from Step 2.

PICTURE. The balance-scale figure: the orange factor $(n + 1)$ on the guess side is cancelled by a plum $\frac{1}{n + 1}$ weight, and the scale settles at $x^{n}$ .

So far we have shown that $\frac{x ^{n + 1}}{n + 1}$ is one function whose slope is $x^{n}$ . In the very next step we will see it is not the only one — every vertical shift of it works too — and that is what forces the extra " $+ C$ " in the boxed rule below. For now, read the rule with a placeholder in mind: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + (some constant, explained in Step 5), n \neq = - 1.$

Step 5 — Where does $+ C$ come from? (the lost-constant picture)

WHAT. Look at $y = x^{2}$ , $y = x^{2} + 3$ , and $y = x^{2} - 2$ . At any fixed $x$ their tangent rulers are parallel — identical slope $2 x$ .

WHY the extra constant is unavoidable. Differentiation flattens a constant to zero: $\frac{d}{d x} (x^{2} + 3) = 2 x, \frac{d}{d x} (x^{2} - 2) = 2 x .$ The $+ 3$ and the $- 2$ leave no trace in the slope. Running backwards, we cannot recover them, so we write a single letter $C$ — the constant of integration — to stand for "some constant I can't know." This is the " $C$ " that appears in every antiderivative.

PICTURE. Three orange copies of the parabola, vertically shifted, with teal tangent rulers at the same $x$ — all pointing the same way. The plum bracket labels the vertical gaps as " $C$ : lost on the way down."

Now the boxed rule earns its final symbol:

Step 6 — The forbidden case: why $n = - 1$ breaks the rule

WHAT. Put $n = - 1$ into the power rule: $\frac{x ^{n + 1}}{n + 1} n = - 1 \frac{x ^{0}}{0} = \frac{1}{0} = undefined.$

$n + 1$ becomes $- 1 + 1 = 0$ .
Dividing by $0$ is meaningless — the whole formula collapses.

WHY it fails geometrically. For $n = - 1$ the integrand is $\frac{1}{x}$ . Raising the power gives $x^{0} = 1$ , whose slope is $0$ , not $\frac{1}{x}$ . The "raise-then-divide" trick literally cannot produce a $\frac{1}{x}$ slope — a different function must.

The patch. The function whose slope is $\frac{1}{x}$ is the natural logarithm (see Natural log and exponential functions). Because $\frac{1}{x}$ also exists for negative $x$ while $ln x$ does not, we use $ln ∣ x ∣$ : $\int \frac{1}{x} d x = ln ∣ x ∣ + C .$

PICTURE. Left panel: the smooth staircase of powers with a hole punched at $n = - 1$ . Right panel: the plum curve $y = ln ∣ x ∣$ plugging that hole — its tangent slope drawn shrinking toward $0$ as $x$ grows, matching $\frac{1}{x}$ .

Step 7 — The hidden fine print: what values of $x$ are even allowed?

WHAT — three kinds of exponent.

$n$ a whole number $\geq 0$ (like $x^{2}, x^{5}$ ): $x^{n}$ is defined for every real $x$ , including negatives and zero. The rule holds for all $x$ .
$n$ a negative whole number $\leq - 2$ (like $x^{- 3}$ ): defined for all $x \neq = 0$ — you cannot divide by zero. The rule holds everywhere except $x = 0$ .
$n$ a fraction / non-integer (like $x^{1/2} = x$ ): here is the subtle part. $x$ asks "what number squared gives $x$ ?" — impossible for $x < 0$ inside the real numbers. So $x^{1/2}$ (and any $x^{n}$ with $n$ non-integer) is only defined for $x \geq 0$ .

WHY this matters for the integral. The identity $\frac{d}{d x} \frac{x ^{n + 1}}{n + 1} = x^{n}$ can only be checked where both sides exist. For a non-integer $n$ that region is $x \geq 0$ (and often $x > 0$ , since at $x = 0$ a negative fractional power blows up). So the honest statement is: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C valid on x > 0 when n is not a whole number.$

Extending to $x < 0$ . For odd roots there is a meaning for negatives — e.g. $3 - 8 = - 2$ , so $x^{1/3}$ is defined for all real $x$ , and the rule extends there. But for even roots ( $x$ , $x^{1/4}$ , …) no real value exists for $x < 0$ , so the antiderivative simply has no real meaning there — you would need complex numbers, which is a different course. The safe habit: when you rewrite a root as a fractional power, immediately note the allowed $x$ .

PICTURE. Three number lines stacked. Top (whole $n \geq 0$ ): the whole line is teal "allowed." Middle (negative whole $n$ ): teal everywhere with a plum hole at $x = 0$ . Bottom (fractional $n$ from an even root): only $x \geq 0$ is teal; $x < 0$ is greyed "no real value."

Step 8 — Two quick worked checks

The tools here extend to composite and product integrands via Integration by substitution and Integration by parts; adding limits turns them into areas through Definite integrals & FTC.

The one-picture summary

Read the whole derivation as a single loop: differentiate drops power & multiplies (orange arrow); integrate raises power & divides (plum arrow); the vertical shift $+ C$ is what the loop cannot recover; the point $n = - 1$ is the one place the loop is broken (patched by $ln ∣ x ∣$ ); and a fractional $n$ quietly restricts the allowed $x$ .

Recall Feynman retelling — say it to a friend

Differentiating a power does two little chores: it slides the exponent to the front as a multiplier, and it knocks the exponent down by one. Integrating is just a detective undoing those two chores in reverse: raise the exponent back up by one, then divide by that new exponent to kill the multiplier that would appear. Because sliding a graph up or down never changes its steepness, the detective can never tell which vertical shift you started from — that unknown shift is the " $+ C$ ." The only spot where the trick self-destructs is $n = - 1$ : raising $- 1$ to $0$ and dividing by $0$ is nonsense, and anyway $x^{0} = 1$ has a flat slope, not a $1/ x$ slope — so nature fills that single gap with $ln ∣ x ∣$ . One last piece of fine print: if the power is a fraction like $\frac{1}{2}$ , then $x^{n}$ only makes sense for $x \geq 0$ (you can't square-root a negative), so the answer only lives on that side of zero.

Connections

4.2.02 Basic integration rules — power, trig, exponential, log (Hinglish) — the parent topic
Derivatives — basic rules — the forward machine we reversed
Integration by substitution · Integration by parts — for composites and products
Definite integrals & FTC — turns these antiderivatives into areas
Natural log and exponential functions — source of the $ln ∣ x ∣$ patch
Trigonometric integrals — the same "read-backwards" idea for trig

Visual walkthrough — Basic integration rules — power, trig, exponential, log