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Integrating xⁿ

Q: What happens when n = −1 in the power rule?

The formula x^n+1n+1 breaks down (division by zero). Integrating x^-1= 1x instead gives |x|+c — a special case covered separately.

Q: How do I find the exact value of c?

You need a specific point (x_1,y_1) that the curve passes through. Substitute it into your general antiderivative (with +c still present) and solve for c.

Every differentiation question has a mirror-image question: given the gradient function, what was the original curve? Integration reverses differentiation exactly — and because it undoes an operation, it always leaves an unknown constant behind.

28 min Video by TLMaths Integration

The big picture

Integration and differentiation are inverse operations, tied together by the Fundamental Theorem of Calculus. If differentiating $x^{n + 1}$ gives $(n + 1) x^{n}$ , then integrating $x^{n}$ must undo that — raise the power by 1 and divide by the new power. But because differentiating ANY constant gives zero, integration can never recover what that constant originally was — so every indefinite integral carries an unknown $+ c$ , representing a whole family of curves that share the same gradient function.

A-Level Maths H1-01: Introducing the Fundamental Theorem of CalculusWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Understand integration as the reverse process of differentiation
Use the rule $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c$ for any $n \neq = - 1$
Integrate a sum of power terms term-by-term
Find $c$ using a given point on the curve

Reversing the power rule

Since differentiating $x^{n + 1}$ gives $(n + 1) x^{n}$ , integrating $x^{n}$ must raise the power by 1 and divide by the new power to undo that scaling exactly.

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c (n \neq = - 1)

Raise the power by 1, divide by the new power, add the constant of integration.

1Raise the power:

x^{4}

2Divide by the new power:

\frac{x ^{4}}{4}

3Add

+ c

Answer

\frac{x ^{4}}{4} + c

Tip — The rule fails at $n = - 1$ (you would divide by zero) — integrating $x^{- 1} = \frac{1}{x}$ instead gives $ln ∣ x ∣ + c$ , covered once logarithmic integration is introduced.

The constant of integration

Because the derivative of ANY constant is zero, reversing differentiation can never tell you what that lost constant originally was. So every indefinite integral must include an arbitrary $+ c$ — representing every possible vertical shift of the antiderivative, all sharing the exact same gradient function.

1Integrate:

y = \int 2 x d x = x^{2} + c

2Substitute the point

(1, 4)

4 = 1^{2} + c

, so

c = 3

Answer

y = x^{2} + 3

Why this works

Without the given point, $y = x^{2} + c$ describes an entire FAMILY of identical-shaped parabolas, each shifted up or down — the extra information (a point on the curve) is what pins down exactly one member of that family.

Integrating sums of powers

Just like differentiation, integration is linear — integrate each term of a polynomial separately, then combine with a single $+ c$ at the end (not one per term).

3 x^{2} \to \frac{3 x ^{3}}{3} = x^{3}

- 4 x \to - \frac{4 x ^{2}}{2} = - 2 x^{2}

5 \to 5 x

(constant terms integrate to

5 x

, since

5 = 5 x^{0}

and

\frac{x ^{1}}{1} = x

Answer

x^{3} - 2 x^{2} + 5 x + c

Tip — Always rewrite roots and fractions as powers of $x$ BEFORE integrating, exactly as you would before differentiating.

Think like an examiner

Never forget the $+ c$ on an indefinite integral — omitting it, even when the algebra is otherwise perfect, typically loses a mark.
Only ONE $+ c$ is needed for a whole sum of terms — do not write a separate constant for each term.
The power rule for integration does not apply at $n = - 1$ ; that special case needs $ln ∣ x ∣$ instead.

Common misconceptions

Omitting the constant of integration

+ c

Every indefinite integral needs a

+ c

, since the reverse of differentiation cannot recover a lost constant term.

Dividing by the OLD power instead of the new one.

After raising the power by 1, divide by that NEW power, e.g.

\int x^{4} d x = \frac{x ^{5}}{5} + c

, not

\frac{x ^{5}}{4}

Forgetting to rewrite

x

\frac{1}{x ^{n}}

as powers before integrating.

Rewrite as

x^{1/2}

x^{- n}

first, then apply the power rule for integration directly.

Reversing the power rule

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c

Valid for n ≠ −1.

\int a d x = a x + c

Integrating a constant.

Stretch yourself

A curve has gradient function $\frac{d y}{d x} = 6 x^{2} - 4 x + 1$ and passes through the point $(- 1, 2)$ . Find the equation of the curve.

Hint — Integrate term-by-term to get a general expression with $+ c$ , then substitute the given point to find the exact value of $c$ .

Questions students ask

Why does integration always need a +c?

Because differentiating any constant term gives zero, the reverse process (integration) has no way to recover what that constant originally was — so every possible constant must be represented by the unknown

+ c

What happens when n = −1 in the power rule?

The formula

\frac{x ^{n + 1}}{n + 1}

breaks down (division by zero). Integrating

x^{- 1} = \frac{1}{x}

instead gives

ln ∣ x ∣ + c

— a special case covered separately.

How do I find the exact value of c?

You need a specific point

(x_{1}, y_{1})

that the curve passes through. Substitute it into your general antiderivative (with

+ c

still present) and solve for

c

Key takeaways

Integration reverses differentiation: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c$ for $n \neq = - 1$ .
Every indefinite integral needs a $+ c$ , since a constant term cannot be recovered.
Integrate polynomials term-by-term, with a single constant at the end.
Use a given point on the curve to find the exact value of $c$ .

How this fits the course

Build on

Implicit and Parametric Differentiation

Definite Integrals

Leads to

Definite Integrals

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